Leetcode-Minimum Window Substrings（unordered_map)

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

For example,
S = "ADOBECODEBANC"
T = "ABC"

Minimum window is "BANC".

Note:
If there is no such window in S that covers all characters in T, return the empty string"".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

算法思想:通过建立hash表unordered_map来记录扫描过的元素，并且记录其重复个数。因为S只能遍历一遍，所以还需要两个指针去限制最终包含区域的范围，即first与behind。且还需要一个count去记录每次循环过程中是否达到了包含字符串T的条件。具体代码如下：

string minWindow(string S, string T) {string result;if (S.empty() || T.empty())return result;unordered_map<char, int> m1; //建立<char,int>类型的hash表，方便记录重复元素unordered_map<char, int> m2;for (int i = 0; i < T.length(); i++)m1[T[i]]++;  //记录target中每个元素的个数int first = 0, behind = 0;//用两个指针去维护字符串Sint count=0;//监控器int minlength = INT_MAX;for (; first < S.length(); first++) {if (m1.find(S[first]) != m1.end()) {m2[S[first]]++;if (m2[S[first]] <= m1[S[first]]) // 元素重复不能比待匹配串多count++;}if (count >= T.length()) {while (m1.find(S[behind]) == m1.end() || m2[S[behind]]>m1[S[behind]]) { //当出现查找不到或者m2的某元素已经多余m1的这个元素时m2[behind]--;behind++; //向后移一位}if (first - behind + 1 < minlength) { //遇到更短的区间时更新minlength = first - behind + 1;result = S.substr(behind, minlength);}}}return result;}