Minimum Window Substring

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 emtpy string "".

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

解题思路:用两个标志位分别表示满足条件的字符串的头尾。尾部索引号确定利用满足当前字符串包含目标字符条件，起始的字符串确定起来比较麻烦，首先若起始的字符为目标字符，且已经搜索到的个数大于目标搜索数，则跳过，其次首字符不是目标字符也跳过；基本按这样的思路写就可以确定符合条件的字符串了，最后通过遍历确定最短的字符串即可。

#include<iostream>#include<vector>#include<string>#include<map>using namespace std;string minWindow(string S, string T) {map<char, int>SubstrMap;        //目标字典（字符，相应个数）map<char, int>BesearchedMap;    //已搜索到的字典for (auto &str : T)SubstrMap[str]++;int CountBesecrsh  = 0;            int minLength      = INT_MAX;int minSubstrBegin = 0;int idx_begin      = 0;for (int idx_end = 0; idx_end != S.size(); ++idx_end){if (SubstrMap.find(S[idx_end]) != SubstrMap.end()){BesearchedMap[S[idx_end]]++;if (BesearchedMap[S[idx_end]]<=SubstrMap[S[idx_end]])CountBesecrsh++;if (CountBesecrsh==T.size())             //表示当前搜索点为尾的字符串满足搜索条件{while (idx_begin<=idx_end){if (SubstrMap.find(S[idx_begin]) == SubstrMap.end()){idx_begin++;continue;}else if (BesearchedMap[S[idx_begin]] > SubstrMap[S[idx_begin]]){BesearchedMap[S[idx_begin]]--;idx_begin++;continue;}elsebreak;}int length = idx_end + 1 - idx_begin;if (length<minLength){minLength      = length;minSubstrBegin = idx_begin;}}}}if (minLength == INT_MAX)return "";return S.substr(minSubstrBegin, minLength);}