leetcode 76 Minimum Window Substring ----- java

Posted 2020-08-14 xiaoba1203

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.

在时间复杂度为O（n)的前提下，求出S中包含T的所有字母的最小字串。

第一次的做法，先用一个256大小的数组记录t中的字母。然后找到每一个符合条件的字符串，比较大小，最后果然超时。

public class Solution {
    public String minWindow(String s, String t) {
        int len1 = s.length(),len2 = t.length();
        int[] tt = new int[256];
        for( int i = 0;i<len2;i++)
            tt[t.charAt(i)]++;
        int[] t2 = tt.clone();
        int num = 0,start = 0,end = 0,length = 0;
        int more = len1;
        for( int i = 0;i<len1;i++){
            if( t2[s.charAt(i)] > 0){
                more = length == 0?len1:length-len2;
                for( int j = i;j<len1;j++){
                    if( t2[s.charAt(j)] > 0){
                        num++;
                        t2[s.charAt(j)]--;
                    }else{
                        more--;
                        if( more == 0)
                            break;
                    }
                    if( num == len2 ){
                        if( length == 0 || (j-i+1) < length ) {
                            start = i;
                            end = j;
                            length = end - start+1;
                            more = length - len2;
                        }
                        t2 = tt.clone();
                        num = 0;
                        break;
                        
                    }
                }    
                t2 = tt.clone();
                num = 0;
            }
            if( length == len2 )
                break;
        }
        if( length == 0 )
            return "";
        char[] result = new char[length];
        for( int i = 0;i<length;i++)
            result[i] = s.charAt(i+start);
        return String.valueOf(result);
    }
}

然后改进一下，两个记录点left，right 都从左到右，right每走一次就在loc数组中记录该字母，并且看此时（left至right）是否包括了所有字母，如果包括，那么找出从left到right之间最小的一组。

速度还算可以。

public class Solution {
    public boolean cover(int[] a,int[] b){
        for( int i = 0;i<256;i++)
            if( a[i] < b[i])
                return false;
        return true;
    }
    
    public String minWindow(String s, String t) {
        int len1 = s.length();
        int len2 = t.length();
        int[] pos = new int[256];
        for( int i = 0;i<len2;i++)
            pos[t.charAt(i)]++;
        int[] loc = new int[256];
        int min = len1+1,start = 0,end = 0;

        for( int left = 0,right = 0;right < len1;right++){
            loc[s.charAt(right)]++;
            if( !cover(loc,pos) )
                continue;

            while( left <= right ){
                char ch = s.charAt(left);
                if( pos[ch] == loc[ch] )
                    break;

                loc[ch]--;
                left++;
            }
            if( right-left < min ){
                min = right-left;
                start = left;
                end = right;
            }
        }
        if( min == len1+1)
            return "";
        char[] result = new char[min+1];
        for( int i = start,j = 0;i<=end;i++,j++)
            result[j] = s.charAt(i);
        return String.valueOf(result);

    }}

 发现每次需要判断一次cover浪费了许多时间，那么加入num变量，记录当前left到right之间，有多少已经命中了的字母。这样速度得到进一步提升。

public class Solution {
    
    
    public String minWindow(String s, String t) {
        int len1 = s.length();
        int len2 = t.length();
        if( len1 == 0 || len2 == 0)
            return "";
        int[] pos = new int[256];
        for( int i = 0;i<len2;i++)
            pos[t.charAt(i)]++;
        int[] loc = new int[256];
        int min = len1+1,start = 0,end = 0,num = 0;
        int left = 0,right = 0;
        while( right < len1 ){
            if( num < len2 ){
                char ch = s.charAt(right);
                if( pos[ch] > 0){
                    loc[ch]++;
                    if( loc[ch] <= pos[ch] )
                        num++;
                }
                right++;
            }
            while( num == len2 ){
                if( min > (right-left)){
                    min = right-left;
                    start = left;
                    end = right;
                }
                char ch = s.charAt(left);
                if( loc[ch] == 0  )
                    left++;
                else if( loc[ch] > pos[ch]){
                    left++;
                    loc[ch]--;
                }else{
                    num--;
                    left++;
                    loc[ch]--;
                }
            }
        }
        if( min == len1+1)
            return "";
        char[] result = new char[min];
        for( int i = start,j = 0;i<end;i++,j++)
            result[j] = s.charAt(i);
        return String.valueOf(result);
    }
}

 还可以在进行一点优化，就是将String转变成char[]这样在数量比较多的时候，会有速度的提升，会基本达到最快

只是加入char[] s_array = s.toCharArray() 以及将 s.charAt(i)变为 s_array[i]。

代码改动不大，就不再是上传一次了。