Leetcode238: Substring with Concatenation of All Words

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本文介绍了一种高效的字符串匹配算法，该算法能在给定的字符串中找到由多个指定长度相同的单词按顺序拼接而成的子串的所有起始位置。通过使用两个哈希映射分别存储目标单词列表及其出现次数和滑动窗口内遇到的单词及其出现次数，实现了快速查找。

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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 wordsexactly once and without any intervening characters.

For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]

You should return the indices: [0,9].

(order does not matter).

class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        int n = words.size();
        vector<int> res;
        if(n<1) return res;
        int l = words[0].size();
        map<string, int> count1;
        int i,j;
        for(i = 0; i < n; i++)
            ++count1[words[i]];
        map<string, int> count2;
        for(i = 0; i <= (int)s.length()-n*l; i++)
        {
            count2.clear();
            for(j = 0; j < n; j++)
            {
                string word = s.substr(i+j*l, l);
                if(count1.find(word) != count1.end())
                {
                    ++count2[word];
                    if(count2[word]>count1[word])
                        break;
                }
                else
                    break;
            }
            if(j == n) 
                res.push_back(i);
        }
        return res;
    }
};