LeetCode 97. Interleaving String

原题链接在这里：https://leetcode.com/problems/interleaving-string/

题目：

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

Example 1:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false

题解：

Let dp[i][j] denotes whether or not s1 till index i and s2 till index j could construct s3 till index i+j.

If current index of s3 pointing to the char is equal to s1 current char, then || dp[i-1][j].

If current index of s3 pointing to the char is equal to s2 current char, then || dp[i][j-1].

Check finally dp[m][n] result.

Note:

current pointing index of s3 is i+j-1. It starts at 1, no need to worry about 0. It is already assigned before.

Time Complexity: O(m*n). m = s1.length(). n = s2.length().

Space: O(m*n).

AC Java:

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        int m = s1.length();
        int n = s2.length();
        if(m+n != s3.length()){
            return false;
        }
        
        boolean [][] dp = new boolean[m+1][n+1];
        dp[0][0] = true;
        for(int j = 1; j<=n; j++){
            if(s3.charAt(j-1) == s2.charAt(j-1)){
                dp[0][j] = true;
            }else{
                break;
            }
        }
        
        for(int i = 1; i<=m; i++){
            if(s3.charAt(i-1) == s1.charAt(i-1)){
                dp[i][0] = true;
            }else{
                break;
            }
        }
        
        for(int i = 1; i<=m; i++){
            for(int j = 1; j<=n; j++){
                char c = s3.charAt(i+j-1);
                if(c == s1.charAt(i-1)){
                    dp[i][j] = dp[i][j] | dp[i-1][j];
                }
                
                if(c == s2.charAt(j-1)){
                    dp[i][j] = dp[i][j] | dp[i][j-1];
                }
            }
        }
        
        return dp[m][n];
    }
}

类似Edit Distance.