97. Interleaving String（字符串的交替连接 动态规划）

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.

 

 

 

 

class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        if len(s1) + len(s2) != len(s3):
            return False
        m=len(s1)
        n=len(s2)
        dp = [[0]*(n+1) for _ in range(m+1)]
        dp[0][0] = 1
        for i in range(1,m+1):
            dp[i][0] = (dp[i-1][0] and s1[i-1] == s3[i+0-1]) 
        for j in range(1,n+1):
            dp[0][j] = (dp[0][j-1] and s2[j-1] == s3[0+j-1])
        for i in range(1,m+1):
            for j in range(1,n+1):
                dp[i][j] = (dp[i-1][j] and s1[i-1] == s3[i+j-1]) or (dp[i][j-1] and s2[j-1] == s3[i+j-1])
        return dp[m][n] == 1

 

动态规划 

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