LeetCode-Substring with Concatenation of All Words-优快云博客

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S中找到由单词列表L构成的所有子串的起始索引，其中L中的所有单词长度相同且每个单词恰好使用一次。

You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.

For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

Solution:

Code:

<span style="font-size:14px;">class Solution {
public:
    vector<int> findSubstring(string S, vector<string> &L) {
        const int lengthS = S.size(), lengthL = L.size();
        const int lengthWord = L[0].size();
        vector<int> results;
        unordered_map<string, int> hashTableL;
        unordered_map<string, int> hashTableS;
        for (int i = 0; i < lengthL; ++i)
            hashTableL[L[i]]++;
        bool succeed;
        for (int i = 0; i <= lengthS-lengthL*lengthWord; ++i) {
            hashTableS.clear();
            succeed = true;
            for (int j = 0; j < lengthL; ++j) {
                string str = S.substr(i+j*lengthWord, lengthWord);
                if (hashTableL.find(str) == hashTableL.end()) {
                    succeed = false;
                    break;
                } else if (++hashTableS[str] > hashTableL[str]) {
                    succeed = false;
                    break;
                }
            }
            if (succeed) results.push_back(i);
        }
        return results;
    }
};</span>