[LeetCode] 30. Substring with Concatenation of All Words
You are given a string s
and an array of strings words
of the same length. Return all starting indices of substring(s) in s
that is a concatenation of each word in words
exactly once, in any order, and without any intervening characters.
You can return the answer in any order.
Example 1:
Input: s = "barfoothefoobarman", words = ["foo","bar"] Output: [0,9] Explanation: Substrings starting at index 0 and 9 are "barfoo" 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: []
Example 3:
Input: s = "barfoofoobarthefoobarman", words = ["bar","foo","the"] Output: [6,9,12]
Constraints:
1 <= s.length <= 104
s
consists of lower-case English letters.1 <= words.length <= 5000
1 <= words[i].length <= 30
words[i]
consists of lower-case English letters.
串联所有单词的子串。
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。
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/substring-with-concatenation-of-all-words
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思路是滑动窗口(sliding window),但是没法套用之前的模板。注意这个题给的条件,words 里面每个单词是等长的,这个条件会使得这个题目简单些。首先创建一个 hashmap 把 words 里面的所有单词和出现次数都记录下来。接着开始遍历字符串 s,每移动一个字符,就要复制一次之前的 hashmap,因为这里的思路是需要看从起点 i 开始的 substring 是否包含 hashmap 里面存的所有单词(及其次数)。当找到一个子串的时候,需要判断这个子串是否还在 hashmap 里,如果这个子串不存在或者次数已经为 0 了则 break,说明以当前 i 为起点的子串无效;如果这个子串存在于 hashmap 则--,同时 words 的个数 K 也要--,这样当 K == 0 的时候就可以知道所有单词都遍历完了,可以把这个子串的起始位置 i 加入结果集了。
时间O(n^2)
空间O(n)
Java实现
1 class Solution { 2 public List<Integer> findSubstring(String s, String[] words) { 3 List<Integer> res = new ArrayList<>(); 4 // corner case 5 if (s == null || words == null || words.length == 0) { 6 return res; 7 } 8 9 // normal case 10 // 单词个数 11 int n = words.length; 12 // 单词长度 13 int m = words[0].length(); 14 HashMap<String, Integer> map = new HashMap<>(); 15 for (String str : words) { 16 map.put(str, map.getOrDefault(str, 0) + 1); 17 } 18 19 for (int i = 0; i <= s.length() - n * m; i++) { 20 HashMap<String, Integer> copy = new HashMap<>(map); 21 int count = n; 22 int j = i; 23 while (count > 0) { 24 String str = s.substring(j, j + m); 25 if (!copy.containsKey(str) || copy.get(str) < 1) { 26 break; 27 } 28 copy.put(str, copy.get(str) - 1); 29 count--; 30 j += m; 31 } 32 if (count == 0) { 33 res.add(i); 34 } 35 } 36 return res; 37 } 38 }
JavaScript实现
1 /** 2 * @param {string} s 3 * @param {string[]} words 4 * @return {number[]} 5 */ 6 var findSubstring = function (s, words) { 7 // corner case 8 if (s == null || words == null || words.length == 0) { 9 return []; 10 } 11 12 // normal case 13 let res = []; 14 let n = words.length; 15 let m = words[0].length; 16 let map = {}; 17 for (let str of words) { 18 map[str] = (map[str] || 0) + 1; 19 } 20 21 for (let i = 0; i <= s.length - n * m; i++) { 22 let copy = { ...map }; 23 let k = n; 24 let j = i; 25 while (k > 0) { 26 let str = s.substring(j, j + m); 27 if (!copy[str] || copy[str] < 1) { 28 break; 29 } 30 copy[str]--; 31 k--; 32 j += m; 33 } 34 if (k == 0) { 35 res.push(i); 36 } 37 } 38 return res; 39 };