我不点开评论都知道评论区有多精彩。这不，有网友立马跳出来辟谣：“你们说错了，Boss直聘也凉了好吗？”这语气有点搞笑，但其实很沉重。
这背后反映的就现在招聘市场的真实情况。但无论如何，对于我们来说，能找到一份满意的工作始终是目标。我们能做的就是提升自己，他强任他强，清风拂山岗。
哈希映射统计：首先使用哈希映射（HashMap）统计words数组中每个单词出现的次数。
滑动窗口：由于所有单词的长度相同，我们可以使用滑动窗口的方式，以单词的长度为步长在原字符串s上滑动，检查每个可能的窗口。
匹配验证：对于每个窗口，使用另一个哈希映射统计窗口中每个单词的出现次数，然后与words的哈希映射进行比较，看是否完全匹配。
记录起始位置：如果一个窗口完全匹配，记录该窗口的起始位置。
优化：为了减少不必要的检查，我们只需要在0到wordLength-1的范围内开始滑动窗口，其中wordLength是数组words中单词的长度。
import java.util.ArrayList;import java.util.HashMap;import java.util.List;import java.util.Map;public class Solution {public List<Integer> findSubstring(String s, String[] words) {int wordLength = words[0].length();int allWordsLength = words.length * wordLength;Map<String, Integer> wordMap = new HashMap<>();List<Integer> result = new ArrayList<>();for (String word : words) {wordMap.put(word, wordMap.getOrDefault(word, 0) + 1);}for (int i = 0; i < wordLength; i++) {for (int j = i; j <= s.length() - allWordsLength; j += wordLength) {Map<String, Integer> seen = new HashMap<>();int k = 0;while (k < words.length) {String word = s.substring(j + k * wordLength, j + (k + 1) * wordLength);if (wordMap.containsKey(word)) {seen.put(word, seen.getOrDefault(word, 0) + 1);if (seen.get(word) > wordMap.get(word)) break;} else {break;}k++;}if (k == words.length) result.add(j);}}return result;}}
function findSubstring(s, words) {const wordLength = words[0].length;const allWordsLength = words.length * wordLength;const wordMap = new Map();const result = [];words.forEach(word => {wordMap.set(word, (wordMap.get(word) || 0) + 1);});for (let i = 0; i < wordLength; i++) {for (let j = i; j <= s.length - allWordsLength; j += wordLength) {const seen = new Map();let k = 0;while (k < words.length) {const word = s.substr(j + k * wordLength, wordLength);if (wordMap.has(word)) {seen.set(word, (seen.get(word) || 0) + 1);if (seen.get(word) > wordMap.get(word)) break;} else {break;}k++;}if (k === words.length) result.push(j);}}return result;}
package mainimport "fmt"func findSubstring(s string, words []string) []int {wordLength := len(words[0])allWordsLength := len(words) * wordLengthwordMap := make(map[string]int)var result []intfor _, word := range words {wordMap[word]++}for i := 0; i < wordLength; i++ {for j := i; j <= len(s)-allWordsLength; j += wordLength {seen := make(map[string]int)var k intfor k = 0; k < len(words); k++ {word := s[j+k*wordLength : j+(k+1)*wordLength]if count, exists := wordMap[word]; exists {seen[word]++if seen[word] > count {break}} else {break}}if k == len(words) {result = append(result, j)}}}return result}
以s = "barfoothefoobarman", words = ["foo","bar"]为例，我们的函数应该返回[0,9]。这是因为在s中，从索引0开始的子串"barfoo"和从索引9开始的子串"foobar"恰好由words中的所有单词串联形成。
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