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.

解法：

双指针，动态维护最小长度，记录最小长度的start 和 end.

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，为指针继续向后扫，遇到在T中的字母时，check头指针是否能收缩，能收缩update最小长度，直到end到达S的end。

 

public class Solution {
    public String minWindow(String S, String T) {
        
        if (S == null || T == null || S.length() == 0 || T.length() == 0) {
            return "";
        }
        int[] needToFind = new int[256];
        int[] hasFound = new int[256];

        for (int i = 0; i < T.length(); i++) {
            needToFind[T.charAt(i)]++;
        }

        int minWinLen = Integer.MAX_VALUE;
        int count = 0, tLen = T.length();
        int winBeg = 0, winEnd = 0;
        for (int begin = 0, end = 0; end < S.length(); end++) {
            if (needToFind[S.charAt(end)] == 0) {
                continue;
            }
            hasFound[S.charAt(end)]++;
            if(hasFound[S.charAt(end)] <= needToFind[S.charAt(end)]){
                count ++;
            }
            
            if(count == tLen){
               //当begin所指的元素不在T中 或  之后又出现了beigin所指的元素 begin向后移动
                while(needToFind[S.charAt(begin)] == 0 || hasFound[S.charAt(begin)] > needToFind[S.charAt(begin)]){
                    if(hasFound[S.charAt(begin)] > needToFind[S.charAt(begin)]){
                        //之后又出现了beigin所指的元素 begin向后移动 重复出现的元素个数减1
                        hasFound[S.charAt(begin)]--;
                    }
                    begin ++;
                }
                
                int winLen = end - begin + 1;
                if(winLen < minWinLen){
                    winBeg = begin;
                    winEnd = end;
                    minWinLen = winLen;
                }
            }
        }

        if (count == T.length()) {
            return S.substring(winBeg, winEnd + 1);
        }

        return "";
    }
}