[LeetCode] 30. 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 "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: []

串联所有单词的子串。题意是给定一个字符串 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。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

class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> res = new ArrayList<>();
        // corner case
        if (s == null || words == null || words.length == 0) {
            return res;
        }

        // normal case
        // 单词个数
        int n = words.length;
        // 单词长度
        int m = words[0].length();
        HashMap<String, Integer> map = new HashMap<>();
        for (String str : words) {
            map.put(str, map.getOrDefault(str, 0) + 1);
        }

        for (int i = 0; i <= s.length() - n * m; i++) {
            HashMap<String, Integer> copy = new HashMap<>(map);
            int count = n;
            int j = i;
            while (count > 0) {
                String str = s.substring(j, j + m);
                if (!copy.containsKey(str) || copy.get(str) < 1) {
                    break;
                }
                copy.put(str, copy.get(str) - 1);
                count--;
                j += m;
            }
            if (count == 0) {
                res.add(i);
            }
        }
        return res;
    }
}

JavaScript实现

/**
 * @param {string} s
 * @param {string[]} words
 * @return {number[]}
 */
var findSubstring = function (s, words) {
    // corner case
    if (s == null || words == null || words.length == 0) {
        return [];
    }

    // normal case
    let res = [];
    let n = words.length;
    let m = words[0].length;
    let map = {};
    for (let str of words) {
        map[str] = (map[str] || 0) + 1;
    }

    for (let i = 0; i <= s.length - n * m; i++) {
        let copy = { ...map };
        let k = n;
        let j = i;
        while (k > 0) {
            let str = s.substring(j, j + m);
            if (!copy[str] || copy[str] < 1) {
                break;
            }
            copy[str]--;
            k--;
            j += m;
        }
        if (k == 0) {
            res.push(i);
        }
    }
    return res;
};

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