leecode更新
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markilue
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ee4c7618f8
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package com.markilue.leecode.hot100;
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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
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*@Author: markilue
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*@CreateTime: 2023-04-06 10:04
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*@Description:
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* TODO 力扣394 字符串解码:
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* 给定一个经过编码的字符串，返回它解码后的字符串。
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* 编码规则为: k[encoded_string]，表示其中方括号内部的 encoded_string 正好重复 k 次。注意 k 保证为正整数。
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* 你可以认为输入字符串总是有效的；输入字符串中没有额外的空格，且输入的方括号总是符合格式要求的。
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* 此外，你可以认为原始数据不包含数字，所有的数字只表示重复的次数 k ，例如不会出现像 3a 或 2[4] 的输入。
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*@Version: 1.0
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*/
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public class T86_DecodeString {
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@Test
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public void test() {
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String s = "3[a]2[bc]";
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System.out.println(decodeString(s));
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}
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@Test
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public void test1() {
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String s = "3[a2[c]]";
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System.out.println(decodeString(s));
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}
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@Test
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public void test2() {
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String s = "3[z]2[2[y]pq4[2[jk]e1[f]]]ef";
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System.out.println(decodeString(s));
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}
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int start = 0;
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public String decodeString(String s) {
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return decodeSubString(s.toCharArray(), 1);
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}
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public String decodeSubString(char[] chars, int count) {
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if (chars[start] == '[') start++;
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StringBuilder sb = new StringBuilder();
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int sum = 0;
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while (start < chars.length && Character.isDigit(chars[start])) {
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sum = sum * 10 + (chars[start] - '0');
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start++;
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}
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//如果前面没有数字,返回1次
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if (sum == 0) {
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while (start < chars.length && chars[start] >= 'a' && chars[start] <= 'z') {
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sb.append(chars[start++]);
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}
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} else {
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sb.append(decodeSubString(chars, sum));
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}
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// if (start < chars.length && Character.isDigit(chars[start])) {
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// sb.append(decodeSubString(chars, 1));
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// }
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if (start < chars.length && chars[start] == ']') {
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start++;
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}
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if (count > 1) {
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String k = sb.toString();
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for (int i = 0; i < count - 1; i++) {
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sb.append(k);
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}
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}
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if (start < chars.length && chars[start] != ']') {
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sb.append(decodeSubString(chars, 1));
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}
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return sb.toString();
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}
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String src;
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int ptr;
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/**
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* 官方递归法:
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* 思路和本人一致，但是思路更清晰和简洁
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* 速度击败100% 内存击败76.28%
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* @param s
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* @return
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*/
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public String decodeString1(String s) {
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src = s;
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ptr = 0;
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return getString();
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}
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public String getString() {
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if (ptr == src.length() || src.charAt(ptr) == ']') {
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// String -> EPS
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return "";
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}
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char cur = src.charAt(ptr);
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int repTime = 1;
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StringBuilder ret = new StringBuilder();
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if (Character.isDigit(cur)) {
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// String -> Digits [ String ] String
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// 解析 Digits
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repTime = getDigits();
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// 过滤左括号
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++ptr;
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// 解析 String
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String str = getString();
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// 过滤右括号
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++ptr;
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// 构造字符串
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while (repTime-- > 0) {
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ret.append(str);
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}
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} else if (Character.isLetter(cur)) {
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// String -> Char String
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// 解析 Char
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ret.append(String.valueOf(src.charAt(ptr++)));
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}
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ret.append(getString());
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return ret.toString();
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}
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public int getDigits() {
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int ret = 0;
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while (ptr < src.length() && Character.isDigit(src.charAt(ptr))) {
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ret = ret * 10 + src.charAt(ptr++) - '0';
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}
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return ret;
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}
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}
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@ -0,0 +1,186 @@
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package com.markilue.leecode.hot100;
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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.hot100
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*@Author: markilue
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*@CreateTime: 2023-04-06 11:47
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*@Description:
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* TODO 力扣399 除法求值:
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* 给你一个变量对数组 equations 和一个实数值数组 values 作为已知条件，其中 equations[i] = [Ai, Bi] 和 values[i] 共同表示等式 Ai / Bi = values[i] 。每个 Ai 或 Bi 是一个表示单个变量的字符串。
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* 另有一些以数组 queries 表示的问题，其中 queries[j] = [Cj, Dj] 表示第 j 个问题，请你根据已知条件找出 Cj / Dj = ? 的结果作为答案。
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* 返回 所有问题的答案 。如果存在某个无法确定的答案，则用 -1.0 替代这个答案。如果问题中出现了给定的已知条件中没有出现的字符串，也需要用 -1.0 替代这个答案。
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* 注意：输入总是有效的。你可以假设除法运算中不会出现除数为 0 的情况，且不存在任何矛盾的结果。
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*@Version: 1.0
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*/
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public class T87_CalcEquation {
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/**
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* 官方题解:并查集解法
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* 速度击败100% 内存击败85.81% 0ms
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* @param equations
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* @param values
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* @param queries
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* @return
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*/
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public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
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int equationSize = equations.size();
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UnionFind unionFind = new UnionFind(2 * equationSize);
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//第一步:预处理，将变量的值与id进行映射，是的并查集的底层使用数组实现，方便编码
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HashMap<String, Integer> hashMap = new HashMap<>(2 * equationSize);
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int id = 0;
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for (int i = 0; i < equationSize; i++) {
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List<String> equation = equations.get(i);
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String var1 = equation.get(0);
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String var2 = equation.get(1);
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if (!hashMap.containsKey(var1)) {
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hashMap.put(var1, id);
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id++;
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}
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if (!hashMap.containsKey(var2)) {
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hashMap.put(var2, id);
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id++;
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}
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unionFind.union(hashMap.get(var1), hashMap.get(var2), values[i]);
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}
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//第二部:做查询
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int queriesSize = queries.size();
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double[] res = new double[queriesSize];
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for (int i = 0; i < queriesSize; i++) {
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String var1 = queries.get(i).get(0);
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String var2 = queries.get(i).get(1);
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Integer id1 = hashMap.get(var1);
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Integer id2 = hashMap.get(var2);
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if (id1 == null || id2 == null) {
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res[i] = -1.0d;
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} else {
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res[i] = unionFind.isConnected(id1, id2);
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}
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}
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return res;
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}
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private class UnionFind {
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private int[] parent;
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private double[] weight;
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public UnionFind(int n) {
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this.parent = new int[n];
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this.weight = new double[n];
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for (int i = 0; i < n; i++) {
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parent[i] = i;
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weight[i] = 1.0d;
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}
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}
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public void union(int x, int y, double value) {
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int rootX = find(x);
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int rootY = find(y);
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if (rootX == rootY) {
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return;
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}
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parent[rootX] = rootY;
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weight[rootX] = weight[y] * value / weight[x];
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}
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/**
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* 路径压缩
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* @param x
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* @return
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*/
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private int find(int x) {
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if (x != parent[x]) {//存在引用链，至少是第一次遍历到才会不相等
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int origin = parent[x];
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parent[x] = find(parent[x]);
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weight[x] *= weight[origin];
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}
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return parent[x];
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}
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public double isConnected(int x, int y) {
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int rootX = find(x);
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int rootY = find(y);
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if (rootY == rootX) {
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return weight[x] / weight[y];
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} else {
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return -1.0d;
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}
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}
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}
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/**
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* 官方DFS法:本质上就是把数据加入子集当中，然后如果存在引用链就一定能够被连接上，则可以计算
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* 速度击败54.5% 内存击败29.67% 1ms
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* @param equations
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* @param values
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* @param queries
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* @return
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*/
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public double[] calcEquation1(List<List<String>> equations, double[] values, List<List<String>> queries) {
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double[] result = new double[queries.size()];
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Map<String, Map<String, Double>> graph = buildGraph(equations, values);
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for (int i = 0; i < queries.size(); i++) {
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//dfs寻找对应的引用
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String start = queries.get(i).get(0);
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String end = queries.get(i).get(1);
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if (!graph.containsKey(start) || !graph.containsKey(end)) {
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result[i] = -1;//有一个没有就肯定计算不出来
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} else {
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Set<String> visited = new HashSet<>();
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result[i] = dfs(graph, start, end, visited);
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}
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}
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return result;
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}
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//构建引用链
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private Map<String, Map<String, Double>> buildGraph(List<List<String>> equations, double[] values) {
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Map<String, Map<String, Double>> graph = new HashMap<>();
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for (int i = 0; i < equations.size(); i++) {
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String v1 = equations.get(i).get(0);
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String v2 = equations.get(i).get(1);
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graph.putIfAbsent(v1, new HashMap<String, Double>());
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graph.get(v1).put(v2, values[i]);
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graph.putIfAbsent(v2, new HashMap<String, Double>());
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graph.get(v2).put(v1, 1.0 / values[i]);
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}
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return graph;
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}
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private double dfs(Map<String, Map<String, Double>> graph, String start, String end, Set<String> visited) {
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visited.add(start);
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Map<String, Double> next = graph.get(start);
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for (Map.Entry<String, Double> entry : next.entrySet()) {
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if (entry.getKey().equals(end)) {
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return entry.getValue();
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}
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if (!visited.contains(entry.getKey())) {
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double nextValue = dfs(graph, entry.getKey(), end, visited);
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if (nextValue > 0) {//即不可能是-1就正常返回
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return entry.getValue() * nextValue;
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}
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}
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}
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return -1;
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}
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}
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