[LeetCode] Minimum Window Substring

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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".

和很多字符串包含的题思路一样，都是维护S字符串的一个窗口，窗口右边一直往后移动，直到窗口内包含所有T中字符串停止，此时窗口左边开始移动，直到窗口刚好不满足包含T中所有字符串则停止，计算更新最小串。

public class Solution {    public String minWindow(String S, String T) {        if (S == null || T == null) {            return "";        }                Map<Character, Integer> map = new HashMap<Character, Integer>();                for (int i = 0; i < T.length(); i++) {            char curChar = T.charAt(i);            if (map.containsKey(curChar)) {                map.put(curChar, map.get(curChar) + 1);            } else {                map.put(curChar, 1);            }        }                String res = "";        int minLen = S.length() + 1;        int pre = 0;        int count = 0;        int len = T.length();                for (int i = 0; i < S.length(); i++) {            char curChar = S.charAt(i);            if (map.containsKey(curChar)) {                map.put(curChar, map.get(curChar) - 1);                if (map.get(curChar) >= 0) {                    count++;                }                                while (count == len) {                    if (map.containsKey(S.charAt(pre))) {                        map.put(S.charAt(pre), map.get(S.charAt(pre)) + 1);                        if (map.get(S.charAt(pre)) > 0) {                            count--;                        }                                                if (i - pre + 1 < minLen) {                            res = S.substring(pre, i + 1);                            minLen = i - pre + 1;                        }                    }                    pre++;                }            }        }                return res;    }}

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