97. Interleaving String

题目：

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.

链接： http://leetcode.com/problems/interleaving-string/

题解：

依然是使用dynamic programming。和Edit Distance很像。假设我们构建一个棋盘，s1代表行，s2代表列，每次只能向右或者向下走，最后看s3这条路径能不能够从左上到达右下。

Time Complexity - O(m * n)， Space Complexity - O(m * n)。

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        if(s1 == null || s2 == null || s3 == null)
            return true;
        int m = s1.length(), n = s2.length(), s = s3.length();
        if(m + n != s)
            return false;
        boolean[][] dp = new boolean[m +1][n + 1];
        dp[0][0] = true;
        
        for(int i = 0; i < m + 1; i++) {
            for(int j = 0; j < n + 1; j++) {
                if(dp[i][j] == true
                || (j - 1 >= 0 && dp[i][j - 1] == true && s2.charAt(j - 1) == s3.charAt(i + j - 1))
                || (i - 1 >= 0 && dp[i - 1][j] == true && s1.charAt(i - 1) == s3.charAt(i + j - 1))) {
                    dp[i][j] = true;
                } else {
                    dp[i][j] = false;
                }
            }
        }
        
        return dp[m][n];
    }
}

还看到有大神用BFS来做，原理其实和DP差不多，复杂度也基本一样，列在参考里了。

题外话: 从9月12号开始，到今天为止，基本leetcode的第一遍就做了100题了，有两道没做，Text Justification和Wild Card Matching，打算学一学Automata以后再来挑战这几道。同时还想学习一下Mining Massive Datasets以及基本的Python编程。目标很多，时间很少，希望一切顺利吧。(10-12-2015)。

Reference：

http://www.cnblogs.com/springfor/p/3896159.html

https://leetcode.com/discuss/19973/8ms-c-solution-using-bfs-with-explanation <- BFS

https://leetcode.com/discuss/22726/dp-solution-in-java

https://leetcode.com/discuss/4667/is-there-a-better-algorithm-than-what-i-give

https://leetcode.com/discuss/11694/my-dp-solution-in-c

https://leetcode.com/discuss/16086/my-solution-in-java-using-dp-time-o-n-m-and-space-o-m