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.

方法一：动态规划，但时间复杂度和空间复杂度太高

class Solution {
public:
    string minWindow(string S, string T) {
        string result;
        if(S.length() == 0 || T.length() == 0) return result;
        if(S.length() < T.length()) return result;
        
        vector<int> layout('z'-'A'+1,0);
        for(int i = 0; i < T.length(); i++)
            layout[T[i]-'A']++;
            
        vector<vector<vector<int>>> record(S.length()+1,vector<vector<int>>(S.length()+1, vector<int>('z'-'A'+1,0)));
        
        for(int i = 1; i < S.length()+1; i++){
            record[i][i][S[i-1]-'A']++;
            if(equal(layout.begin(),layout.end(),record[i][i].begin())){
                result = S.substr(i-1,1);
                return result;
            }
        }
        for(int l = 2; l <= T.length(); l++){
            for(int i = 1; i < S.length()+2-l; i++){
                record[i][i+l-1] = record[i][i+l-2];
                record[i][i+l-1][S[i+l-2]-'A']++;
                if(equal(layout.begin(),layout.end(),record[i][i+l-1].begin())){
                    result = S.substr(i-1,l);
                    return result;
                }
            }    
        }
        return result;
    }
};

方法二：双指针，先移动win_end指针直到包含整个string T，再移动win_start指针缩小范围至一个极小window。继续上述操作，每次找到一个极小window，如果其小于当前最小window，更新最小window，直至win_end到达S末尾。

class Solution {
public:
    string minWindow(string S, string T) {
        string result;
        if(S.length() == 0 || T.length() == 0) return result;
        if(S.length() < T.length()) return result;
        
        vector<int> expected_count(256,0);
        vector<int> appeared_count(256,0);
        
        for(int i = 0; i < T.length(); i++)
            expected_count[T[i]]++;
        
        int win_start = 0, win_end = 0;
        int min_win_size = INT_MAX, min_win_start = 0;
        int appeared = 0;
        
        for(win_end = 0; win_end < S.length(); win_end++){
            if(expected_count[S[win_end]] > 0){
                appeared_count[S[win_end]]++;
                if(appeared_count[S[win_end]] <= expected_count[S[win_end]])
                    appeared++;
            }
            if(appeared == T.length()){
                while(appeared_count[S[win_start]] > expected_count[S[win_start]] || expected_count[S[win_start]] == 0){
                    appeared_count[S[win_start]]--;
                    win_start++;
                }
                if(min_win_size > (win_end - win_start + 1)){
                    min_win_size = win_end - win_start + 1;
                    min_win_start = win_start;
                }
            }
        }
        
        if(min_win_size == INT_MAX) return "";
        return S.substr(min_win_start,min_win_size);
    }
};