标签:
Notes:
1. When check left shifting, do not use continue, but break it!!! Otherwise, you wil fall into infinite loop
2. Initialize the map with T, not S!!!!
1 class Solution { 2 public: 3 string minWindow(string S, string T) { 4 int len = S.size(), count = 0, length = INT_MAX; 5 string result; 6 if (len == 0 || len < T.size()) return ""; 7 unordered_map<char, int> dict; 8 for (char c : T) { 9 if (dict[c]) dict[c]++; 10 else dict[c] = 1; 11 } 12 for (int left = 0, right = 0; right < len; right++) { 13 if (dict.find(S[right]) != dict.end()) { 14 dict[S[right]]--; 15 if (dict[S[right]] == 0) { 16 count++; 17 } 18 } 19 20 while (left < len) { 21 if (dict.find(S[left]) != dict.end() && dict[S[left]] >= 0) break; 22 if (dict.find(S[left]) != dict.end()) { 23 dict[S[left]]++; 24 } 25 left++; 26 } 27 28 if (count == dict.size() && length > (right - left + 1)) { 29 result = S.substr(left, right-left+1); 30 length = right - left + 1; 31 } 32 } 33 return result; 34 } 35 };
LeetCode – Refresh – Minimum Window Substring
标签:
原文地址:http://www.cnblogs.com/shuashuashua/p/4355111.html