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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).
思路:是一道维护一个动态窗口的题目,维护一个[start, end)的窗口来包含L中的串,每次移动的单位都是len(L的串都是同样大小的)
class Solution { public: vector<int> findSubstring(string S, vector<string> &L) { map<string, int> wordTimes; for (int i = 0; i < L.size(); i++) { if (wordTimes.count(L[i]) == 0) wordTimes[L[i]] = 1; else wordTimes[L[i]]++; } int len = L[0].size(); vector<int> res; for (int i = 0; i < len; i++) { map<string, int> curTimes; int start = i, cnt = 0; for (int end = i; end <= (int)S.size()-len; end += len) { string word = S.substr(end, len); if (wordTimes.find(word) != wordTimes.end()) { if (curTimes.find(word) == curTimes.end()) curTimes[word] = 1; else curTimes[word]++; if (curTimes[word] <= wordTimes[word]) cnt++; else { for (int k = start; ; k += len) { string tmp = S.substr(k, len); curTimes[tmp]--; if (tmp == word) { start = k + len; break; } cnt--; } } if (cnt == L.size()) res.push_back(start); } else { start = end + len; curTimes.clear(); cnt = 0; } } } return res; } };
LeetCode Substring with Concatenation of All Words
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原文地址:http://blog.csdn.net/u011345136/article/details/43918199