[LeetCode 解題報告]030. 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: []

考察：字符串查找，題目意思：給定字符串s，words是一個字符串數組 ，每個字符串的長度是一樣。這裏我們設words數組長度爲m，每個字符串的長度爲n，求這m個字符串拼接成m*n的大字符串在S出現的index。

使用unordered_map 保存words字符串出現的次數，進行哈希映射判斷；然後枚舉s的每個index作爲起始位置，切割字符串進行查找判斷；

class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        if (s.empty() || words.empty() || s.length() < words.size()*words[0].size())
            return {};
        vector<int> res;
        unordered_map<string, int> m1;
        for (int i = 0; i < words.size(); i ++)
            m1[words[i]] ++;
        
        int m = words.size(), n = words[0].size();
        for (int i = 0; i <= s.length()-m*n; i ++) {
            unordered_map<string, int> m2;
            int j;
            for (j = 0; j < m; j ++) {
                string seg = s.substr(i+j*n, n); //切割
                if (m1.find(seg) == m1.end())  //是否在m1出現，沒出現，則此i不符合
                    break;
                m2[seg] ++;
                if (m1[seg] < m2[seg]) // 出現次數多了，則此i不符合
                    break;
            }
            if (j == m) // 此i符合
                res.push_back(i);
        }
        return res;
       
    }
};

完，