LeetCode 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.

解题思路
典型序列型动态规划，Two Sequence DP，Time Complexity - O(m * n)， Space Complexity - O(m * n)。

方式一：dfs。boolean[][] f 代表f[i][j]，是否被遍历过。如果被遍历过，就不遍历了，直接回退，继续别的搜索。

Runtime: 1 ms runtime beats 93.42% of java submission.

	public boolean isInterleave(String a, String b, String c) {
		if (a.length() + b.length() != c.length()) return false;
		return isInterleave(0, 0, new boolean[a.length() + 1][b.length() + 1], a, b, c);
	}

	public boolean isInterleave(int i, int j, boolean[][] f, String a, String b, String c) {
		if (f[i][j]) return false;
		else f[i][j] = true;
		if (i == a.length() && j == b.length())
			return true;
		if (i < a.length() && a.charAt(i) == c.charAt(i + j) && isInterleave(i + 1, j, f, a, b, c))
			return true;
		if (j < b.length() && b.charAt(j) == c.charAt(i + j) && isInterleave(i, j + 1, f, a, b, c))
			return true;
		return false;
	}

方式二： 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。

Runtime: 6 ms runtime beats 48.44% of java submissions.

	public boolean isInterleave2(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];
	}