LeetCode刷题：30. Substring with Concatenation of All Words
原题链接：https://leetcode.com/problems/substring-with-concatenation-of-all-words/
You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
Example 1:
Input:
s = "barfoothefoobarman",
words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.
Example 2:
Input:
s = "wordgoodgoodgoodbestword",
words = ["word","good","best","word"]
Output: []
给定一个字符串 s 和一些长度相同的单词 words。找出 s 中恰好可以由 words 中所有单词串联形成的子串的起始位置。
注意子串要与 words 中的单词完全匹配，中间不能有其他字符，但不需要考虑 words 中单词串联的顺序。
示例 1：
输入：
s = "barfoothefoobarman",
words = ["foo","bar"]
输出：[0,9]
解释：
从索引 0 和 9 开始的子串分别是 "barfoor" 和 "foobar" 。
输出的顺序不重要, [9,0] 也是有效答案。
示例 2：
输入：
s = "wordgoodgoodgoodbestword",
words = ["word","good","best","word"]
输出：[]
算法设计
package com.bean.algorithmbasic;import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;/** Substring with Concatenation of All Words* */
public class LeetCode_30 {public List<Integer> findSubstring(String s, String[] words) {List<Integer> result = new ArrayList<>();int wordSize = words.length;if (wordSize == 0) {return result;}// words 数组中每个字符的长度都一致, 取第一个int wordLength = words[0].length();// 字符数组有可能有重复的, 使用 allWordMap 存放每个 word 出现的次数HashMap<String, Integer> allWordMap = new HashMap<>();for (String word : words) {allWordMap.put(word, allWordMap.getOrDefault(word, 0) + 1);}for (int i = 0; i < s.length() - wordSize * wordLength + 1; i++) {HashMap<String, Integer> hashMap = new HashMap<>();// 单趟循环中, 成功匹配的次数int num = 0;while (num < wordSize) {// 根据查找字符的长度进行截断String word = s.substring(i + num * wordLength, i + (num + 1) * wordLength);if (allWordMap.containsKey(word)) {hashMap.put(word, hashMap.getOrDefault(word, 0) + 1);// 还有可能超过 words 中原来的数量if (hashMap.get(word) > allWordMap.get(word)) {break;}} else {// 如果没有查询到, 跳过这次循环, 查找下一个字符break;}num++;}if (num == wordSize) {result.add(i);}}return result;}public static void main(String[] args) {// TODO Auto-generated method stub/** Input: s = "barfoothefoobarman", words = ["foo","bar"] Output: [0,9]* Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar"* respectively. The output order does not matter, returning [9,0] is fine too.*/LeetCode_30 leetCode30 = new LeetCode_30();String s="barfoothefoobarman";String[] words= {"foo","bar"};List<Integer> list = leetCode30.findSubstring(s, words);System.out.println(list);}}
程序运行结果：
[0, 9]