LeetCode – Refresh – Substring with Concatenation of All

For this problem:

1. use i = start; i < start + total; i += each and s.substr(i, each)

2. or i = 0; i < len; i++ and s.substr(start + i*each, each);

 1 class Solution {
 2 public:
 3     bool findExist(string s, int start, int each, int len, unordered_map<string, int> dict) {
 4         for (int i = 0; i < len; i++) {
 5             if (dict[s.substr(start + i*each, each)] > 0) dict[s.substr(start + i*each, each)]--;
 6             else return false;
 7         }
 8         return true;
 9     }
10     vector<int> findSubstring(string S, vector<string> &L) {
11         vector<int> result;
12         if (L.size() == 0) return result;
13         int len = L.size(), each = L[0].size(), total = each*len, i = 0;
14         unordered_map<string, int> dict;
15         for (string s : L) {
16             dict[s]++;
17         }
18         while (i + total <= S.size()) {
19             if (findExist(S, i, each, len, dict)) {
20                 result.push_back(i);
21             }
22             i++;
23         }
24         return result;
25     }
26 };

posted on 2015-03-24 15:37 keepshuatishuati 阅读(109) 评论(0) 编辑收藏举报