[LeetCode] 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.

» Solve this problem

[解题报告]
双指针，动态维护一个区间。尾指针不断往后扫，当扫到有一个窗口包含了所有T的字符后，然后再收缩头指针，直到不能再收缩为止。最后记录所有可能的情况中窗口最小的

class Solution {public:    string minWindow(string S, string T) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int sLen=S.size();        int tLen=T.size();        if(tLen==0 || sLen<tLen) return "";                int needFind[256]={0};        int hasFind[256]={0};                for(int i=0;i<tLen;i++)        {            needFind[T[i]]++;        }                int minWindowLength=INT_MAX;        int minBegin=0;        int minEnd=sLen-1;        int begin=0;        int end=0;        for(int count=0;end<sLen;end++)        {            //skip those not in T            if(needFind[S[end]]==0) continue;                        hasFind[S[end]]++;            //update the total number of characters found(aaa only counts 2 for aa in T)            if(hasFind[S[end]] <= needFind[S[end]])                count++;                            //a window exists from begin to end            if(count==tLen)            {                //move begin pointer to find the minimum window                while(begin<end)                {                    if(needFind[S[begin]]==0)                     {                        begin++;                        continue;                    }                    if(hasFind[S[begin]] > needFind[S[begin]])                    {                        hasFind[S[begin]]--;                        begin++;                        continue;                    }                    else                        break;                }                                int tmpWindowLength=end-begin+1;                                if(tmpWindowLength < minWindowLength)                {                    minBegin=begin;                    minEnd=end;                    minWindowLength=tmpWindowLength;                }            }        }                if(minWindowLength==INT_MAX)            return "";        return S.substr(minBegin,minWindowLength);    }};