LeetCode Interleaving String(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.

题意：给出字符串s1,s2,s3，问s3是否能用s1和s2的元素交叉构成

思路：用f(i,j)表示用字符串s1(1..i)和s2(1..j)是否可以组成 s3(1..i+j)

如果s1(i)==s3(i+j),有f(i,j)=f(i - 1,j)

如果s2(j)==s3(i+j),有f(i,j)=f(i,j-1)

具体代码如下：

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