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).
这题真是自作孽不可活啊。一直想用动态规划去做，但是一旦L里面有重复词就很麻烦。花了几个小时还是没做出来啊。。。我可怜的周末。
网上一看，原来就是扫描每一个位置，从该位置开始对每个单词计数。map里面存的是每个单词出现的次数（可能大于1）。
注意点：
1. S.length()返回值是size_t，如果直接用S.length()-int的话，负数会被转成很大的数；所以以后记得取字符串大小时，用一个int先来存。(Line 9)；
2. 用一个map统计word list中每个单词出现的次数； (Line 10-13)
3. 从第i个位置开始，用一个map来统计每个单词出现的次数，该次数不能超过第2步得到的次数；(Line 19)；
4. 如果最终合法的单词总数等于L.size()的话，把i存到结果中；（Line 25)。
1 class Solution { 2 public: 3 vector<int> findSubstring(string S, vector<string> &L) { 4 vector<int> ret; 5 if (S.empty()) return ret; 6 if (L.empty()) return ret; 7 int len = L[0].size(); 8 int wholeLen = len * L.size(); 9 int n = S.length(); 10 map<string, int> wordCount; 11 for (int j = 0; j < L.size(); ++j) { 12 wordCount[L[j]]++; 13 } 14 for (int i = 0; i <= n - wholeLen; ++i) { 15 string tmp = S.substr(i, len); 16 int pos = i; 17 int c = 0; 18 map<string, int> exist; 19 while (wordCount.count(tmp) > 0 && exist[tmp] < wordCount[tmp]) { 20 c++; 21 pos += len; 22 exist[tmp]++; 23 tmp = S.substr(pos, len); 24 } 25 if (c == L.size()) ret.push_back(i); 26 } 27 return ret; 28 } 29 30 private: 31 32 };
Leetcode | Substring with Concatenation of All Words,布布扣,bubuko.com
Leetcode | Substring with Concatenation of All Words
原文：http://www.cnblogs.com/linyx/p/3722054.html