Interleaving String (DP)

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.

State:f[i][j]  表示s1的前i 个字符 和 s2的前j 个字符能组成s3的前 i + j 个字符

Function: if (((s1.charAt(i - 1) == s3.charAt(i + j - 1) && interleave[i - 1][j])) || (s2.charAt(j - 1) == s3.charAt(i + j - 1) && interleave[i][j - 1])) {
                    interleave[i][j] = true;

Initializtion:f[0][0] = true    f[i][0]  i = (1 ~ s1.length() )  f[0][j]  j = (1 ~ s2.length())

Answer:f[s1.length()][s2.length()]

 public class Solution {
     public boolean isInterleave(String s1, String s2, String s3) {
         if (s1.length() + s2.length() != s3.length()) {
             return false;
         }

         boolean[][] interleave = new boolean[s1.length() + 1][s2.length()  + 1];
         interleave[0][0] = true;
         for (int i = 1; i <= s1.length(); i++) {
             if (s1.charAt(i - 1) == s3.charAt(i - 1) && interleave[i - 1][0]) {
                 interleave[i][0] = true;
             }
         }
         for (int i = 1; i <= s2.length(); i++) {
             if (s2.charAt(i - 1) == s3.charAt(i - 1) && interleave[0][i - 1]) {
                 interleave[0][i] = true;
             }
         }

         for (int i = 1; i <= s1.length(); i++) {
             for (int j = 1; j <= s2.length(); j++) {
                 if (((s1.charAt(i - 1) == s3.charAt(i + j - 1) && interleave[i - 1][j])) || (s2.charAt(j - 1) == s3.charAt(i + j - 1) && interleave[i][j - 1])) {
                     interleave[i][j] = true;
                 }
             }
         }
         return interleave[s1.length()][s2.length()];
     }
 }