leecode更新
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markilue
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5317d93f39
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1104f87ef2
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@ -0,0 +1,207 @@
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package com.markilue.leecode.backtrace;
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import org.junit.Test;
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import java.util.*;
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/**
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*@BelongsProject: Leecode
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*@BelongsPackage: com.markilue.leecode.backtrace
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*@Author: dingjiawen
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*@CreateTime: 2023-02-03 10:21
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*@Description:
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* TODO 力扣332 重新安排行程:
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* 给你一份航线列表 tickets ，其中 tickets[i] = [fromi, toi] 表示飞机出发和降落的机场地点。请你对该行程进行重新规划排序。
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* 所有这些机票都属于一个从 JFK（肯尼迪国际机场）出发的先生，所以该行程必须从 JFK 开始。如果存在多种有效的行程，请你按字典排序返回最小的行程组合。
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* 例如，行程 ["JFK", "LGA"] 与 ["JFK", "LGB"] 相比就更小，排序更靠前。
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* 假定所有机票至少存在一种合理的行程。且所有的机票 必须都用一次 且 只能用一次。
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*@Version: 1.0
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*/
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public class T13_0_FindItinerary {
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@Test
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public void test() {
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List<List<String>> tickets = new ArrayList<>();
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tickets.add(Arrays.asList("MUC", "LHR"));
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tickets.add(Arrays.asList("JFK", "MUC"));
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tickets.add(Arrays.asList("SFO", "SJC"));
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tickets.add(Arrays.asList("LHR", "SFO"));
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findItinerary(tickets);
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System.out.println(result);
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}
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@Test
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public void test1() {
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List<List<String>> tickets = new ArrayList<>();
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tickets.add(Arrays.asList("JFK", "SFO"));
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tickets.add(Arrays.asList("JFK", "ATL"));
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tickets.add(Arrays.asList("SFO", "ATL"));
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tickets.add(Arrays.asList("ATL", "JFK"));
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tickets.add(Arrays.asList("ATL", "SFO"));
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findItinerary1(tickets);
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System.out.println(result);
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}
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@Test
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public void test2() {
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List<List<String>> tickets = new ArrayList<>();
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tickets.add(Arrays.asList("JFK", "KUL"));
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tickets.add(Arrays.asList("JFK", "NRT"));
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tickets.add(Arrays.asList("NRT", "JFK"));
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// tickets.add(Arrays.asList("ATL", "JFK"));
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// tickets.add(Arrays.asList("ATL", "SFO"));
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findItinerary1(tickets);
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System.out.println(result);
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}
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List<String> cur = new ArrayList<>();
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List<String> result = new ArrayList<>();
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List<List<String>> result1 = new ArrayList<>();
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public List<String> findItinerary(List<List<String>> tickets) {
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Collections.sort(tickets, new Comparator<List<String>>() {
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@Override
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public int compare(List<String> o1, List<String> o2) {
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return o1.get(1).compareTo(o2.get(0));
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}
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});
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backtracking(tickets, 0, "", new boolean[tickets.size()]);
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return result;
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}
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public void backtracking(List<List<String>> tickets, int level, String last, boolean[] used) {
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if (level == tickets.size()) {
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//所有机票用完了
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result = new ArrayList<>(cur);
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return;
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}
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for (int i = 0; i < tickets.size(); i++) {
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if (level == 0) {
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if (!"JFK".equals(tickets.get(i).get(0))) {
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//起始地一定要是JFK
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continue;
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}
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used[i] = true;
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List<String> lines = tickets.get(i);
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cur.add(lines.get(0));
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cur.add(lines.get(1));
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backtracking(tickets, level + 1, lines.get(1), used);
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if (result != null) {
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return;
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}
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used[i] = false;
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cur.clear();
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} else {
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if (used[i] || !last.equals(tickets.get(i).get(0))) {
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continue;
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}
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used[i] = true;
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cur.add(tickets.get(i).get(1));
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backtracking(tickets, level + 1, tickets.get(i).get(1), used);
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if (result != null) {
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return;
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}
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used[i] = false;
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cur.remove(cur.size() - 1);
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}
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}
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}
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/**
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* 代码优化
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* 速度击败26.88%，内存击败43.82% 15ms
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* @param tickets
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* @return
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*/
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public List<String> findItinerary1(List<List<String>> tickets) {
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Collections.sort(tickets, (a, b) -> a.get(1).compareTo(b.get(1)));
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cur.add("JFK");
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backtracking1(tickets, 0, "JFK", new boolean[tickets.size()]);
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return result;
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}
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public void backtracking1(List<List<String>> tickets, int level, String last, boolean[] used) {
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if (level == tickets.size()) {
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//所有机票用完了
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result = new ArrayList<>(cur);
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return;
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}
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for (int i = 0; i < tickets.size(); i++) {
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if (used[i] || !last.equals(tickets.get(i).get(0))) {
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continue;
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}
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used[i] = true;
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cur.add(tickets.get(i).get(1));
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backtracking1(tickets, level + 1, tickets.get(i).get(1), used);
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if (result != null && result.size() != 0) {
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return;
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}
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used[i] = false;
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cur.remove(cur.size() - 1);
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}
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}
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/**
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* 官方解法:使用一个优先队列进行排序，好像默认了按字典序就一定能遍历完所有的票
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* 速度击败98.38%，内存击败5.4% 5ms
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*/
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Map<String, PriorityQueue<String>> map = new HashMap<String, PriorityQueue<String>>();
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List<String> itinerary = new LinkedList<String>();
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public List<String> findItinerary2(List<List<String>> tickets) {
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for (List<String> ticket : tickets) {
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String src = ticket.get(0), dst = ticket.get(1);
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if (!map.containsKey(src)) {
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map.put(src, new PriorityQueue<String>());
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}
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map.get(src).offer(dst);
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}
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dfs("JFK");
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Collections.reverse(itinerary);
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return itinerary;
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}
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public void dfs(String curr) {
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while (map.containsKey(curr) && map.get(curr).size() > 0) {
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String tmp = map.get(curr).poll();
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dfs(tmp);
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}
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itinerary.add(curr);
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}
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/**
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* 官方最快 4ms,不用反转了，进一步优化
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*/
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private List<String> resList = new LinkedList<>();
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public List<String> findItinerary3(List<List<String>> tickets) {
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for (List<String> ticket : tickets) {
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String src = ticket.get(0);
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String dst = ticket.get(1);
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if (!map.containsKey(src)) {
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PriorityQueue<String> pq = new PriorityQueue<>();
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map.put(src, pq);
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}
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map.get(src).add(dst);
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}
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dfs1("JFK");
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return resList;
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}
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private void dfs1(String src) {
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PriorityQueue<String> pq = map.get(src);
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while (pq != null && !pq.isEmpty()){
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dfs1(pq.poll());
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}
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((LinkedList<String>) resList).addFirst(src);
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}
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}
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@ -17,7 +17,7 @@ import java.util.*;
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* 每一种解法包含一个不同的 n 皇后问题 的棋子放置方案，该方案中 'Q' 和 '.' 分别代表了皇后和空位。
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* @Version: 1.0
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*/
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public class SolveNQueens {
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public class T13_1_SolveNQueens {
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@Test
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public void test() {
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@ -0,0 +1,201 @@
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package com.markilue.leecode.backtrace.second;
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import org.junit.Test;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.List;
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/**
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*@BelongsProject: Leecode
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*@BelongsPackage: com.markilue.leecode.backtrace.second
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*@Author: dingjiawen
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*@CreateTime: 2023-02-03 12:43
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*@Description:
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* TODO 力扣51题 N皇后:
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* 按照国际象棋的规则，皇后可以攻击与之处在同一行或同一列或同一斜线上的棋子。
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* n 皇后问题 研究的是如何将 n 个皇后放置在 n×n 的棋盘上，并且使皇后彼此之间不能相互攻击。
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* 给你一个整数 n ，返回所有不同的 n 皇后问题 的解决方案。
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* 每一种解法包含一个不同的 n 皇后问题 的棋子放置方案，该方案中 'Q' 和 '.' 分别代表了皇后和空位。
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*@Version: 1.0
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*/
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public class T13_1_SolveNQueens {
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@Test
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public void test() {
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System.out.println(solveNQueens1(4));
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}
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List<List<String>> result = new ArrayList<>();
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List<String> cur = new ArrayList<>();
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public List<List<String>> solveNQueens(int n) {
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backtracking(n, 0, new boolean[n][n]);
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return result;
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}
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/**
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* 速度击败41.44%，内存击败78.99% 4ms
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* @param n
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* @param level
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* @param used
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*/
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public void backtracking(int n, int level, boolean[][] used) {
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if (level == n) {
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for (int i = 0; i < n; i++) {
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StringBuilder builder = new StringBuilder();
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for (int j = 0; j < n; j++) {
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if(used[i][j]){
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builder.append("Q");
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}else {
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builder.append(".");
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}
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}
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cur.add(builder.toString());
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}
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result.add(new ArrayList<>(cur));
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cur.clear();
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return;
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}
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StringBuilder builder = new StringBuilder();
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for (int i = 0; i < n; i++) {
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if (isFit(used, level, i)) {
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used[level][i] = true;
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backtracking(n, level + 1, used);
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used[level][i] = false;
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}
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}
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}
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public boolean isFit(boolean[][] used, int row, int col) {
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//看看列
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for (int i = 0; i < row; i++) {
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if (used[i][col]) {
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return false;
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}
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}
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//看看行,行不用看，一定不会
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//看看斜边
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for (int i = row - 1, j = col - 1, k = col + 1; i >= 0; i--, j--, k++) {
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if (j >= 0 && used[i][j]) {
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return false;
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}
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if (k < used.length && used[i][k]) {
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return false;
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}
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}
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return true;
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}
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/**
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* 代码随想录式，不用一个一个append速度提高
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* 速度击败92.55%，内存击败48.62% 2ms
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*/
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List<List<String>> res = new ArrayList<>();
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public List<List<String>> solveNQueens1(int n) {
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char[][] chessboard = new char[n][n];
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for (char[] c : chessboard) {
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Arrays.fill(c, '.');
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}
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backTrack(n, 0, chessboard);
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return res;
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}
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public void backTrack(int n, int row, char[][] chessboard) {
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if (row == n) {
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res.add(Array2List(chessboard));
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return;
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}
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for (int col = 0;col < n; ++col) {
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if (isFitWithChar(chessboard,row, col)) {
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chessboard[row][col] = 'Q';
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backTrack(n, row+1, chessboard);
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chessboard[row][col] = '.';
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}
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}
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}
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public List Array2List(char[][] chessboard) {
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List<String> list = new ArrayList<>();
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for (char[] c : chessboard) {
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list.add(String.copyValueOf(c));
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}
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return list;
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}
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public boolean isFitWithChar(char[][] used, int row, int col) {
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//看看列
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for (int i = 0; i < row; i++) {
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if (used[i][col]=='Q') {
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return false;
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}
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}
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//看看行,行不用看，一定不会
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//看看斜边
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for (int i = row - 1, j = col - 1, k = col + 1; i >= 0; i--, j--, k++) {
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if (j >= 0 && used[i][j]=='Q') {
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return false;
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}
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if (k < used.length && used[i][k]=='Q') {
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return false;
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}
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}
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return true;
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}
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/**
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* 官方最快 0ms
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* 通过大量的位运算，和Arrays.fill加快速度
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* @param n
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* @return
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*/
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public List<List<String>> solveNQueens2(int n) {
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List<List<String>> ans=new ArrayList<>();
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int[] queen=new int[n];
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Arrays.fill(queen,-1);
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backtrack(ans,n,queen,0,0,0,0);
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return ans;
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}
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public void backtrack(List<List<String>> ans,int n,int[] queen,int rows,int col,int diagonal1,int diagonal2){
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if(rows==n){
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List<String> list=new ArrayList<>();
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for(int i=0;i<n;i++){
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char[] row=new char[n];
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Arrays.fill(row,'.');
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row[queen[i]]='Q';
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list.add(new String(row));
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}
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ans.add(list);
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}else{
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int available=((1<<n)-1) & (~(col | diagonal1 | diagonal2));
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while(available!=0){
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int position=available&(-available);
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available=available & (available-1);
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queen[rows]=Integer.bitCount(position-1);
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backtrack(ans,n,queen,rows+1,position|col,(position|diagonal1)<<1,(position|diagonal2)>>1);
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queen[rows]=-1;
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}
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}
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}
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}
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