[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.

https://oj.leetcode.com/problems/minimum-window-substring/

思路：窗口移动法。 详见这里

public class Solution {
    public String minWindow(String S, String T) {
        if (S == null || T == null)
            return "";

        int[] needToFind = new int[256];
        int[] found = new int[256];

        for (int i = 0; i < T.length(); i++)
            needToFind[T.charAt(i)]++;
        String result = "";
        int begin = 0, end = 0;
        int minWin = Integer.MAX_VALUE;
        int count = 0;

        for (end = 0; end < S.length(); end++) {
            if (needToFind[S.charAt(end)] == 0)
                continue;

            char ch = S.charAt(end);
            found[ch]++;
            if (found[ch] <= needToFind[ch])
                count++;
            
            if (count == T.length()) {
                //move the begin pointer while the constrain meets
                while (begin < S.length()
                        && (found[S.charAt(begin)] > needToFind[S.charAt(begin)] || needToFind[S.charAt(begin)] == 0)) {

                    if (found[S.charAt(begin)] > 0)
                        found[S.charAt(begin)]--;

                    begin++;
                }

                //update the min according to the current window
                if (end - begin + 1 < minWin) {
                    minWin = end - begin + 1;
                    result = S.substring(begin, end + 1);
                }

            }

        }
        return result;

    }

    public static void main(String[] args) {
        System.out.println(new Solution().minWindow("ADOBECODEBANC", "ABC"));

    }

}

第二遍记录：

　　注意T中可能有重复元素，所以needToFind和found都要用int[]

　　注意begin指针移动的条件，同时记得更新found数组。

第四遍记录：

　　注意count值从找到第一个窗口之后就不变了，一直等于T.length，然后之后移动end，每次遇到T中字符就再次尝试移动左窗口。