[leetcode]Minimum Window Substring

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.

算法思路：窗口移动法。 详见这里

本题算法略吊，还是看详解吧。

代码如下：

public class Solution {
    public String minWindow(String s, String t) {
        if(s == null || t == null) return "";
        int[] found = new int[256];
        int[] needToFind = new int[256];
        for(int i = 0; i < t.length(); i++){
            needToFind[t.charAt(i)]++;
        }
        int start = 0, count = 0,minWin = Integer.MAX_VALUE;
        String res = "";
        for(int i = 0; i < s.length(); i++){
            if(needToFind[s.charAt(i)] == 0) continue;
            char c = s.charAt(i);
            found[c]++;
            if(found[c] <= needToFind[c])
                count++;
            if(count == t.length()){
                //move the begin pointer while the constrain meets
                while(start < s.length() && ( needToFind[s.charAt(start)] == 0 || found[s.charAt(start)] > needToFind[s.charAt(start)])){
                    if(found[s.charAt(start)] > 0)
                        found[s.charAt(start)]--;
                    start++;
                }
                //update the min according to the current window
                if(i - start + 1 < minWin){
                    minWin = i - start + 1;
                    res = s.substring(start,i + 1);
                }
            }
        }
        return res;
    }
}

 

第二遍：

minWin 初始化最好 > s.length()，否则当窗口大小== s.length时候，如果有重复最小窗口时候，代码可能就会稍微麻烦一点了。

未通过case："abc", "ac"

这是一道很好的题目

 

参考：

http://www.cnblogs.com/jdflyfly/p/3815275.html