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.

思路：

1. 维护一个map<Character, Integer>来表示每个字母还可以出现的次数。开始的时候建立map，字母对应在target string里出现的次数。另外一个count

2. 对长sring扫描，如果遇到map里存在的字母，就对对应的值-1 （注意这里可以是负数，表示该位字母出现多次，但是我们不计算在总count里）.如果该值>=0则总count++，表示总共找到的字母增加一位。

3. 如果count = target.length()说明已经全部找到，这时窗口左边向右移动，如果对应的值在map里，那就把map里的value+1，因为我们需要寻找的字母又少了一个。在value+1后，如果count<target.length()，说明我们又可以开始寻找下一个了，就先计算一下当前match的长度：right-left+1

    public String minWindow(String S, String T) {        if (T.length() == 0) {            return "";        }        int count = 0;        Map<Character, Integer> map = new HashMap<Character, Integer>();        for (int i = 0; i < T.length(); i++) {            char c = T.charAt(i);            if (map.get(c) == null) {                map.put(c, 1);            } else {                map.put(c, map.get(c)+1);            }        }        int minLen = Integer.MAX_VALUE;        String result = "";        int left = 0, right = 0;        while (left <= S.length() - T.length() && right < S.length()) {            if (map.containsKey(S.charAt(right))) {                map.put(S.charAt(right), map.get(S.charAt(right))-1);                if (map.get(S.charAt(right)) >= 0) {                    count++;                }                while (count == T.length()) {                    if (map.get(S.charAt(left)) != null) {                        map.put(S.charAt(left), map.get(S.charAt(left))+1);                        if (map.get(S.charAt(left)) > 0) {                            int curLen = right-left+1;                            if (curLen < minLen) {                                minLen = Math.min(minLen, curLen);                                result = S.substring(left, right+1);                            }                            count--;                        }                    }                    left++;                }            }            right++;        }        return result;    }