LeetCode-Interleaving String[dp]

Interleaving String

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.

 

标签: Dynamic Programming String

分析：动态规划；设boolean数组dp[i][j]表示字符串s1[0...i-1]和字符串s2[0...j-1]是否可以匹配到字符串s3[0...i+j-1],

因此有字符s3[i+j-1]可以由字符s1[i-1]或字符s2[j-1]来匹配，所以：

如果s1[i-1]==s3[i+j-1],则dp[i][j]=dp[i-1][j];

如果s2[j-1]==s3[i+j-1],则dp[i][j]=dp[i][j-1];

所以状态转移方程为：dp[i][j]=(dp[i-1][j]&&s1[i-1]==s3[i+j-1])||(dp[i][j-1]&&s2[j-1]==s3[i+j-1])

参考代码：

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
            int len1=s1.length();
            int len2=s2.length();
            int len3=s3.length();
            if(len1+len2!=len3)
                return false;
            boolean dp[][]=new boolean[len1+1][len2+1];
            dp[0][0]=true;
            for(int j=1;j<=len2;j++){
                dp[0][j]=dp[0][j-1]&&s2.charAt(j-1)==s3.charAt(j-1);
            }
            for(int i=1;i<=len1;i++){
                dp[i][0]=dp[i-1][0]&&s1.charAt(i-1)==s3.charAt(i-1);
            }
            for(int i=1;i<=len1;i++){
                for(int j=1;j<=len2;j++){
                    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[len1][len2];
    }
}