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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).
class Solution { public: vector<int> findSubstring(string S, vector<string> &L) { vector<int> res; int len=L[0].size()*L.size(); if (S.empty()||L.empty()||S.size()<len)return res; map<string,int> Lmap; map<string,int> temp; for (int i=0;i<L.size();i++) Lmap[L[i]]++; for (int i=0;i<=(int)S.size()-len;i++) { temp.clear(); int j=i; for (;j<i+len;j+=L[0].size()) { string word=S.substr(j,L[0].size()); if (Lmap.find(word)==Lmap.end()) break; temp[word]++; if (temp[word]>Lmap[word]) break; } if (j==i+len) res.push_back(i); } return res; } };
leetcode[30]Substring with Concatenation of All Words
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原文地址：http://www.cnblogs.com/Vae98Scilence/p/4283617.html
