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 empty string"".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

Solution: 类似于All anagrams。因为window长度不定，所以用2指针。j一直增加，直到match == hashmap.size()，更新min和start和end，然后i++，并更新hashmap，直到不match。

    public String minWindow(String s, String t) {
        if (s.length() < t.length() || t.length() == 0 || s.length() == 0) {
            return "";
        }
        Map<Character, Integer> hashmap = new HashMap<>();
        for (int i = 0; i < t.length(); i++) {
            char c = t.charAt(i);
            hashmap.put(c, hashmap.getOrDefault(c, 0) + 1);
        }
        int i = 0, j = 0, match = 0, min = Integer.MAX_VALUE, start = -1, end = -1;
        while (j < s.length()) {
            //要用while而不是if continue，因为有可能j == s.length()时match了
            while (j < s.length() && match < hashmap.size()) {
                char c = s.charAt(j);
                Integer count = hashmap.get(c); 
                if (count != null) {
                    hashmap.put(c, count - 1);
                    //count从1变为0时，说明个数凑满，match++
                    if (count == 1) {
                        match++;
                    }
                }
                j++;
            }
            while (match == hashmap.size()) {
                if (j - i < min) {
                    min = j - i;
                    start = i;
                    end = j;
                }            
                char c = s.charAt(i);
                Integer count = hashmap.get(c); 
                if (count != null) {
                    hashmap.put(c, count + 1);
                    //count从0变为1时，说明个数从凑满变为缺一个，match--
                    if (count == 0) {
                        match--;
                    }
                }
                i++;
            }
        }
        if (start == -1) {
            return "";
        }
        return s.substring(start, end);
    }