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.

Solution:

Code:

<span style="font-size:14px;">class Solution {
public:
    string minWindow(string S, string T) {
        int lengthS = S.size(), lengthT = T.size();
        if (lengthT > lengthS) return "";
        unordered_map<char, int> hashTable;
        for (int i = 0; i < lengthT; i++)
            hashTable[T[i]]++;
        int count = 0, begin = 0, end = 0;
        for (; end < lengthS; end++)
            if (hashTable.find(S[end]) != hashTable.end()) {
                hashTable[S[end]]--;
                if (hashTable[S[end]] >= 0) count++;
                if (count == lengthT) break;
            }
        if (count != lengthT) return "";
        for (; begin <= end; begin++)
            if (hashTable.find(S[begin]) == hashTable.end()) continue;
            else if (hashTable[S[begin]] < 0) hashTable[S[begin]]++;
            else if (hashTable[S[begin]] == 0) break;
        int minLength = end-begin+1;
        int result = begin;
        end++;
        for (; end < lengthS; end++)
            if (S[begin] == S[end]) {
                begin++;
                for (; begin <= end; begin++)
                    if (hashTable.find(S[begin]) == hashTable.end()) continue;
                    else if (hashTable[S[begin]] < 0) hashTable[S[begin]]++;
                    else if (hashTable[S[begin]] == 0) break;
                if (end-begin+1 < minLength) {
                    minLength = end-begin+1;
                    result = begin;
                }
            } else {
                if (hashTable.find(S[end]) != hashTable.end())
                    hashTable[S[end]]--;
            }
        return S.substr(result, minLength);
    }
};</span>