97. 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.

设状态为 f[i][j]，表示 S3 的 0-(i+j+1)的子串是 S1 的 0-i 子串和 S2 的 0-j
子串的 interleaving string
则状态转移方程为：

f[i][j] = True  f[i-1][j] && S1[i-1]==S3[i+j-1] 
                    or   f[i][j-1] && S2[j-1]==S3[i+j-1]
        = False else

	null	d	b	b	c	a
null	T	F	F	F	F	F
a	F	F	F	F	F	F
a	T	T	T	T	T	F
b	F	T	T	F	F	F
c	F	F	T	T	T	T
c	F	F	F	T	F	T

s3 = "aadbbcbcac"
使用滚动数组优化空间复杂度

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