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.

动态规划。
dp[i][j]代表s1的前i个字符，s2的前i个字符是否能构造出s3的i+j个字符。如果dp[i-1][j]为真，则只需要比较s1的第i个字符，同s3的i+j个字符；如果dp[i][j-1]为真，则只需要比较s2的第j个字符，同s3的i+j个字符。代码如下

 public boolean isInterleave(String s1, String s2, String s3) {
        if(s1 == null || s1.length() == 0) return s2.equals(s3);
        if(s2 == null || s2.length() == 0) return s1.equals(s3);
        if(s3 == null || s1.length() + s2.length() != s3.length()) return false;
        int l1 = s1.length(),l2 =  s2.length();
        boolean[][] dp = new boolean[l1+1][l2+1];
        dp[0][0] = true;
        for (int i = 0; i <= l1; i++) {
            for (int j = 0; j <= l2; j++) {
                if (i == 0 && j == 0) {
                    continue;
                }
                if (i-1 >= 0 && dp[i-1][j] == true && s1.charAt(i-1) == s3.charAt(i+j-1)) {
                    dp[i][j] = true;
                    continue;
                }
                if (j-1 >= 0 && dp[i][j-1] == true && s2.charAt(j-1) == s3.charAt(i+j-1)) {
                    dp[i][j] = true;
                }
            }
        }
        return dp[l1][l2];
    }