leetcode 97 Interleaving String ----- java

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

For example,
Given:
s1 = "aabcc",
s2 = "dbbca",

When s3 = "aadbbcbcac", return true.
When s3 = "aadbbbaccc", return false.

 

给定三个字符串，判断第三个字符串是否由前两个字符串组合而成（互相插入，即相对顺序不变。）

1、第一次写的是贪心算法，但是超时了。

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
		
		
		if( s1.length()+s2.length() != s3.length())
			return false;
			
		if( s1.length()+s2.length() != s3.length())
			return false;
		if( s1.length() == 0)
			return s2.equals(s3);
		if( s2.length() == 0)
			return s1.equals(s3);
		System.out.println(s1.length()+" "+s2.length()+" "+s3.length() );
		return getResult(s1,s2,s3,0,0,0);
        
    }

    public boolean getResult(String s1,String s2,String s3,int i,int j,int k){
    	if( i == s1.length() && j == s2.length() && k == s3.length())
    		return true;
    	else if( i == s1.length() ){
    		while( j<s2.length() && s2.charAt(j) == s3.charAt(k)){
    			j++;
    			k++;
    		}
    		if( j == s2.length() )
    			return true;
    		else 
    			return false;
    	}else if( j == s2.length() ){
    		while( i<s1.length() && s1.charAt(i) == s3.charAt(k)){
    			i++;
    			k++;
    		}
    		if( i == s1.length() )
    			return true;
    		else 
    			return false;
    	}else if( s1.charAt(i) == s3.charAt(k) && s2.charAt(j) != s3.charAt(k))
    		return getResult(s1,s2,s3,i+1,j,k+1);
    	else if( s2.charAt(j) == s3.charAt(k) && s1.charAt(i) != s3.charAt(k))
    		return getResult(s1,s2,s3,i,j+1,k+1);
    	else if( s1.charAt(i) == s3.charAt(k) && s2.charAt(j) == s3.charAt(k)){
    		boolean res = getResult(s1,s2,s3,i+1,j,k+1);
    		if( res )
    			return true;
    		res = getResult(s1,s2,s3,i,j+1,k+1);
    		return res?true:false;
    	}else 
    		return false;




    }
}

 2、所以使用DP算法

这里就类似于一个二维的地图，判断是否能从左上角走到右下角。

示例：s1="aa",s2="ab",s3="aaba"。标1的为可行。最终返回右下角。

     0  a  b

0   1  1  0

a   1  1  1

a   1  0  1

 

最后AC，但是并不算太快。

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        
        if( s1.length()+s2.length() != s3.length())
            return false;
        if( s1.length() == 0)
            return s2.equals(s3);
        if( s2.length() == 0)
            return s1.equals(s3);
        boolean[][] result = new boolean[s1.length()+1][s2.length()+1];
        result[0][0] = true;
        for( int i = 0;i<=s1.length();i++){
            for( int j = 0;j<=s2.length();j++){
                if( i == 0 && j == 0)
                    result[i][j] = true;
                else if( j == 0){
                    result[i][0] = result[i-1][0] && s1.charAt(i-1) == s3.charAt(i-1);
                }else if( i == 0){
                    result[0][j] = result[0][j-1] && s2.charAt(j-1) == s3.charAt(j-1);
                }
                else{
                    if( result[i-1][j]  && s1.charAt(i-1) == s3.charAt(i+j-1))
                        result[i][j] = true;
                    else if( result[i][j-1] && s2.charAt(j-1) == s3.charAt(i+j-1))
                        result[i][j] = true;
                }

            }
        }
        return result[s1.length()][s2.length()];
    }
}

3、可以在空间上稍微简化一下，使用单数组完成。只是稍作改变，就不再贴出来了。

4、第三种方法是将上述两种方法结合一下，可以达到1ms，最快，这里是借鉴了Discuss里面的一个答案。

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        int[][] memo = new int[s1.length() + 1][s2.length() + 1];
        return helper(s1, s2, s3, 0, 0, 0, memo);
    }
    
    private boolean helper(String s1, String s2, String s3, int index1, int index2, int index3, int[][] memo) {
        if(s1.length() + s2.length() != s3.length()) return false;
        if(s1.length() == index1 && s2.length() == index2  && s3.length() == index3 ) return true;

        boolean s1flag = false;
        boolean s2flag = false;
        
        if(index1 < s1.length() && s1.charAt(index1) == s3.charAt(index3)) {
            if(memo[index1+1][index2] == 1) {
                s1flag = true;
            } else if(memo[index1][index2] == 0){
                s1flag = helper(s1, s2, s3, index1+1, index2, index3+1, memo);
                memo[index1+1][index2] = s1flag ? 1 : -1;
            }
        }
        
        if(index2 < s2.length() && s2.charAt(index2) == s3.charAt(index3)) {
            if(memo[index1][index2+1] == 1) {
                s2flag = true;
            } else if(memo[index1][index2] == 0){
                s2flag = helper(s1, s2, s3, index1, index2 + 1, index3 + 1, memo);
                memo[index1][index2+1] = s1flag ? 1 : -1;
            }
        }
        
        return s1flag || s2flag;
    }
}