[解题思路]

DP

Let F(i, j) denote if s1 of length i and s2 of length j could form s3 of lenght i+j, then we have four cases:
(1) F(i, j) = F(i-1, j)                              if s1[i-1] == s3[i+j-1] && s2[j-1] != s3[i +j -1]
(2) F(i, j) = F(i, j-1)                              if s1[i-1] != s3[i+j-1]  && s2[j-1] == s3[i +j -1]
(3) F(i, j) = F(i-1, j) || F(i, j-1)             if s1[i-1] == s3[i+j-1] && s2[j-1] == s3[i +j -1]
(4) F(i, j) = false                                   if s1[i-1] != s3[i+j-1] && s2[j-1] != s3[i +j -1]

但是merge sort没法考虑两个字符串的组合顺序问题。当处理{"C","CA", "CAC"}的时候，就不行了。

最后还是得用DP。对于
s1 = a1, a2 ........a(i-1), ai
s2 = b1, b2, .......b(j-1), bj
s3 = c1, c3, .......c(i+j-1), c(i+j)

定义 match[i][j] 意味着，S1的(0, i)和S2的(0,j)，匹配与S3的(i+j)
如果 ai == c(i+j), 那么 match[i][j] = match[i-1][j]， 等价于如下字符串是否匹配。

s1 = a1, a2 ........a(i-1)
s2 = b1, b2, .......b(j-1), bj
s3 = c1, c3, .......c(i+j-1)

同理，如果bj = c(i+j), 那么match[i][j] = match[i][j-1];

所以，转移方程如下：

Match[i][j]
    =   (s3.lastChar == s1.lastChar) && Match[i-1][j]
      ||(s3.lastChar == s2.lastChar) && Match[i][j-1]
初始条件：
    i=0 && j=0时，Match[0][0] = true;
    i=0时， s3[j] = s2[j], Match[0][j] |= Match[0][j-1]
           s3[j] != s2[j], Match[0][j] = false;

    j=0时， s3[i] = s1[i], Match[i][0] |= Match[i-1][0]
           s3[i] != s1[i], Match[i][0] = false;

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int len1 = s1.length(), len2 = s2.length(), len3 = s3.length();
        if(len3 != len1 + len2){
            return false;
        }
        
        boolean[][] match = new boolean[len1 + 1][len2 + 1];
        match[0][0] = true;
        
        // j == 0
        int i = 1;
        while(i <= len1 && s1.charAt(i-1) == s3.charAt(i-1)){
            match[i][0] = true;
            i++;
        }
        
        // i == 0
        int j = 1;
        while(j <= len2 && s2.charAt(j-1) == s3.charAt(j-1)){
            match[0][j] = true;
            j++;
        }
        
        for(int m = 1; m <= len1; m++){
            for(int n = 1; n <= len2; n++){
                char c = s3.charAt(m + n - 1);
                if(c == s1.charAt(m - 1)){
                    match[m][n] |= match[m-1][n];
                }
                if(c == s2.charAt(n - 1)){
                    match[m][n] |= match[m][n-1];
                }
            }
        }
        return match[len1][len2];
    }
}

ref: http://www.cnblogs.com/feiling/p/3296057.html   && http://fisherlei.blogspot.com/2012/12/leetcode-interleaving-string.html

[Leetcode]--Interleaving String