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.

两个指针+字符hash

string minWindow(string S, string T) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int nT = T.size();        int nS = S.size();                int needToFind[256] = {0};        for (int i = 0; i < nT; ++i)            ++needToFind[T[i]];        int hasFound[256] = {0};        int minBegin;        int minEnd;        int minWindow = nS + 1;        int count = 0;                char ch;        for (int begin = 0, end = 0; end < nS; ++end)        {            if (needToFind[S[end]] == 0)                continue;            ch = S[end];            ++hasFound[ch];            if (hasFound[ch] <= needToFind[ch])                ++count;                            if (count == nT)            {                while (needToFind[S[begin]] == 0                || hasFound[S[begin]] > needToFind[S[begin]])                {                    if (hasFound[S[begin]] > needToFind[S[begin]])                        --hasFound[S[begin]];                    ++begin;                }                                int length = end - begin + 1;                if (length < minWindow)                  {                    minBegin = begin;                    minEnd = end;                    minWindow = length;                }            }        }                return minWindow <= nS ? S.substr(minBegin, minEnd-minBegin+1) : "";    }