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.

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        int length_s1 = s1.length();
        int length_s2 = s2.length();
        int length_s3 = s3.length();
        if(length_s1 + length_s2 != length_s3)
            return false;
        boolean state[][] = new boolean[length_s1 + 1][length_s2 + 1];
        //init state[][]
        state[0][0] = true;
        for(int i = 1; i <= length_s1; i++){
            state[i][0] = (s1.charAt(i - 1) == s3.charAt(i - 1)) && state[i - 1][0];
        }//for
        for(int i = 1; i <= length_s2; i++){
            state[0][i] = (s2.charAt(i - 1) == s3.charAt(i - 1)) && state[0][i - 1];
        }//for
        
        //fill state[][]
        for(int i = 1; i <= length_s1; i++){
            for(int j = 1; j <= length_s2; j++){                
                state[i][j] = (state[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1)) || (state[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1));
            }//for
        }//for
        //Util.showTwoDimensionalArray(state);
        return state[length_s1][length_s2];
    }
}

思路

这是一道二维动态规划的题

参考：http://blog.csdn.net/sunbaigui/article/details/8980830

这里state[i][j]表示s3(1..i + j )个字符是否能由s1(1, i)和s2（1， j）个字符串组成

递推公式

state[i][j] = (state[i - 1][j] &&　s1.charAt(i - 1) )|| (state[i][j - 1] && s2.charAt(j - 1))

如果用递归会出现超时的问题