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: []

给一组子串，再给一个主串，问主串中是否存在子串组的任意排列组合。

首先枚举起始位置，然后枚举该位置后面的子串是否能满足给定子串组的排列组合，具体是用 unordered_map 来完成（内部用散列表实现）

class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        unordered_map<string, int> c;
        vector<int> res;
        int m=words.size();
        for(int i=0;i<m;i++)
            c[words[i]]++;
        int n=s.length();
        if(n==0||m==0)return res;
        int len=words[0].length();
        for(int i=0;i<n-m*len+1;i++){
            unordered_map<string, int> vis;
            int j=0;
            for(;j<m;j++){
                string w=s.substr(i+j*len,len);
                if(c.find(w)!=c.end()){
                    vis[w]++;
                    if(vis[w]>c[w])break;
                }else break;
            }
            if(j==m)res.push_back(i);
        }
        return res;
    }
};