leecode更新
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markilue
parent
1f12cec9f0
commit
47beed6aa1
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@ -1,5 +1,7 @@
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package com.markilue.leecode.hot100.interviewHot.difference;
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import org.junit.Test;
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/**
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*@BelongsProject: Leecode
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*@BelongsPackage: com.markilue.leecode.hot100.interviewHot.difference
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@ -15,11 +17,59 @@ package com.markilue.leecode.hot100.interviewHot.difference;
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*/
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public class LC_1094_CarPooling {
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@Test
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public void test() {
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int[][] trips = {{2, 1, 5}, {
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3, 5, 7
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}};
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int capacity = 4;
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System.out.println(carPooling(trips, capacity));
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}
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/**
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* 差分数组:本质上就是获得每个位置的一个大小，判断是否超过capacity
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* 而数组又是范围型累加，所以可以使用差分数组法
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* 速度击败99.33% 内存击败5.5% 1ms
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* @param trips
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* @param capacity
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* @return
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*/
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public boolean carPooling(int[][] trips, int capacity) {
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//构造差分数组
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int max = Integer.MIN_VALUE;
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int min = Integer.MAX_VALUE;
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for (int[] trip : trips) {
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if (trip[1] < min) min = trip[1];
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if (trip[2] > max) max = trip[2];
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}
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//通过最大值最小值构建差分数组
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int[] diff = new int[max - min + 1];
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for (int[] trip : trips) {
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diff[trip[1] - min] += trip[0];
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int end = trip[2] - min;//当前就需要减掉
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if (end < diff.length) {
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diff[end] -= trip[0];
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}
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}
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if (diff[0] > capacity) {
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return false;
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}
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//挨个遍历判断是否合适
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for (int i = 1; i < diff.length; i++) {
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diff[i] += diff[i - 1];
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if (diff[i] > capacity) {
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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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@ -103,4 +103,32 @@ public class LC_1109_CorpFlightBookings {
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}
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return nums;
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}
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/**
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* 二刷:差分
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* @param bookings
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* @param n
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* @return
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*/
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public int[] corpFlightBookings3(int[][] bookings, int n) {
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//构建差分数组
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int[] diff = new int[n];
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for (int[] booking : bookings) {
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diff[booking[0] - 1] += booking[2];
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if (booking[1] < n) {
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diff[booking[1]] -= booking[2];
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}
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}
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//通过差分数组还原当前位置
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for (int i = 1; i < diff.length; i++) {
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diff[i] += diff[i - 1];
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}
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return diff;
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}
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}
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@ -25,7 +25,7 @@ public class T35_76_MinWindow {
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@Test
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public void test1() {
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String s = "ADOBECODEBANC", t = "ABC";
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System.out.println(minWindow1(s, t));
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System.out.println(minWindow2(s, t));
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}
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@ -131,4 +131,55 @@ public class T35_76_MinWindow {
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return result == Integer.MAX_VALUE ? "" : s.substring(start, start + result);
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}
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public String minWindow2(String s, String t) {
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HashMap<Character, Integer> need = new HashMap<>();
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//遍历t,获取每个字母需要的个数
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for (int i = 0; i < t.length(); i++) {
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char c = t.charAt(i);
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need.put(c, need.getOrDefault(c, 0) + 1);
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}
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HashMap<Character, Integer> window = new HashMap<>();//窗口
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int left = 0;
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int right = 0;
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int diff = need.size();
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int minLength = Integer.MAX_VALUE;
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int start = 0;
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while (right < s.length()) {
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char c = s.charAt(right++);
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if (need.containsKey(c)) {
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Integer count = window.getOrDefault(c, 0);
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if (count.equals(need.get(c) - 1)) {
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diff--;
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}
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window.put(c, count + 1);
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}
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while (diff == 0 && left < right) {
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int curLength = right - left;//由于之前right++了,所以这里不用+1了
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if (curLength < minLength) {
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minLength = curLength;
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start = left;
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}
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char c1 = s.charAt(left++);
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if (need.containsKey(c1)) {
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Integer count = window.getOrDefault(c1, 0);
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if (count.equals(need.get(c1))) {
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diff++;
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}
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window.put(c1, count - 1);
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}
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}
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}
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return minLength == Integer.MAX_VALUE ? "" : s.substring(start, start + minLength);
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}
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}
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@ -0,0 +1,89 @@
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package com.markilue.leecode.hot100.second;
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import com.markilue.leecode.tree.TreeNode;
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import java.util.ArrayList;
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import java.util.LinkedList;
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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.hot100.second
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*@Author: markilue
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*@CreateTime: 2023-05-06 12:51
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*@Description:
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* TODO 力扣94 二叉树的中序遍历:
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* 给定一个二叉树的根节点 root ，返回 它的 中序 遍历 。
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*@Version: 1.0
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*/
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public class T40_94_InorderTraversal {
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//递归法
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public List<Integer> inorderTraversal(TreeNode root) {
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ArrayList<Integer> list = new ArrayList<>();
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inorderTraversal(root, list);
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return list;
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}
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public void inorderTraversal(TreeNode root, List<Integer> list) {
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if (root == null) {
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return;
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}
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inorderTraversal(root.left, list);
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list.add(root.val);
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inorderTraversal(root.right, list);
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}
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//morris遍历法
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public List<Integer> inorderTraversal1(TreeNode root) {
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ArrayList<Integer> list = new ArrayList<>();
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while (root != null) {
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if (root.left != null) {
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TreeNode left = root.left;
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while (left.right != null && left.right != root) {
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left = left.right;
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}
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//判断是怎么出来的
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if (left.right == null) {
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left.right = root;
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//放心将root左移
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root = root.left;
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} else {
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//第二次到这里了
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list.add(root.val);
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root = root.right;
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left.right = null;
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}
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} else {
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list.add(root.val);
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root = root.right;
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}
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}
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return list;
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}
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//stack栈法
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public List<Integer> inorderTraversal2(TreeNode root) {
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ArrayList<Integer> result = new ArrayList<>();
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LinkedList<TreeNode> stack = new LinkedList<>();
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while (!stack.isEmpty() || root != null) {
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if (root != null) {
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stack.push(root);
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root = root.left;
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} else {
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TreeNode cur = stack.pop();
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result.add(cur.val);
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root = cur.right;
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}
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}
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return result;
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}
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}
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@ -0,0 +1,35 @@
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package com.markilue.leecode.hot100.second;
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import org.junit.Test;
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/**
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*@BelongsProject: Leecode
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*@BelongsPackage: com.markilue.leecode.hot100.second
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*@Author: markilue
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*@CreateTime: 2023-05-06 13:18
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*@Description:
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* TODO 力扣96 不同的二叉搜索树:
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* 给你一个整数 n ，求恰由 n 个节点组成且节点值从 1 到 n 互不相同的 二叉搜索树 有多少种？返回满足题意的二叉搜索树的种数。
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*@Version: 1.0
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*/
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public class T41_96_NumTrees {
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public int numTrees(int n) {
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if (n <= 2) {
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return n;
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}
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int[] dp = new int[n + 1];
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dp[1] = 1;
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dp[0] = 1;
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for (int i = 2; i < dp.length; i++) {
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for (int j = 0; j < i; j++) {
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dp[i] += dp[j] * dp[i - j - 1];
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}
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}
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return dp[n];
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}
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}
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@ -0,0 +1,46 @@
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<?xml version="1.0" encoding="UTF-8"?>
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<project xmlns="http://maven.apache.org/POM/4.0.0"
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xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
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xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
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<modelVersion>4.0.0</modelVersion>
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<groupId>com.markilue.interview</groupId>
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<artifactId>MeiTuan</artifactId>
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<version>1.0-SNAPSHOT</version>
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<properties>
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<maven.compiler.source>8</maven.compiler.source>
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<maven.compiler.target>8</maven.compiler.target>
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</properties>
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<dependencies>
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<dependency>
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<groupId>junit</groupId>
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<artifactId>junit</artifactId>
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<version>4.13.2</version>
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<scope>test</scope>
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</dependency>
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<dependency>
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<groupId>junit</groupId>
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<artifactId>junit</artifactId>
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<version>4.13.2</version>
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<scope>compile</scope>
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</dependency>
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<dependency>
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<groupId>org.projectlombok</groupId>
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<artifactId>lombok</artifactId>
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<version>RELEASE</version>
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<scope>compile</scope>
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</dependency>
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</dependencies>
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</project>
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@ -32,7 +32,7 @@ public class Question2 {
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int k = 3;
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int[] nums = {1, 2, 3, 2, 1, 4, 5, 1};
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sovle(k,nums);
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solve(k, nums);
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}
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@ -78,4 +78,41 @@ public class Question2 {
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}
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public void solve(int k, int[] nums) {
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HashMap<Integer, Integer> window = new HashMap<>();//count
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int left = 0;
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int right = 0;
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int count = 0;
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int maxLength = Integer.MIN_VALUE;
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while (right < nums.length) {
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int num = nums[right++];
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if (!window.containsKey(num)) {
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count++;
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}
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window.put(num, window.getOrDefault(num, 0) + 1);
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if (count == k) {
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int curLength = right - left;
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if (curLength > maxLength) {
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maxLength = curLength;
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}
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}
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while (count > k && left < right) {
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int num1 = nums[left++];
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int numLeft = window.getOrDefault(num1, 0) - 1;
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if (numLeft == 0) {
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window.remove(num1);
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count--;
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}
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}
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}
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System.out.println(maxLength);
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}
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}
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