CODE 46. Scramble String

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版权声明：本文为博主原创文章，转载请注明出处：http://blog.jerkybible.com/2013/09/21/2013-09-21-CODE 46 Scramble String/

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Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.
Below is one possible representation of s1 = "great":
great
/ \
gr eat
/ \ / \
g r e at
/ \
a t

To scramble the string, we may choose any non-leaf node and swap its two children.
For example, if we choose the node "gr" and
swap its two children, it produces a scrambled string "rgeat".
rgeat
/ \
rg eat
/ \ / \
r g e at
/ \
a t

We say that "rgeat" is
a scrambled string of "great".
Similarly, if we continue to swap the children of nodes "eat" and "at",
it produces a scrambled string "rgtae".
rgtae
/ \
rg tae
/ \ / \
r g ta e
/ \
t a

We say that "rgtae" is
a scrambled string of "great".
Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

public boolean isScramble(String s1, String s2) {
	// Start typing your Java solution below
	// DO NOT write main() function
	if (null == s1 && null == s2) {
		return true;
	} else if ((null == s1 && null != s2) || (null != s1 && null == s2)) {
		return false;
	} else if (s1.length() != s2.length()) {
		return false;
	} else if (s1.equals(s2)) {
		return true;
	}
	char[] c1 = s1.toCharArray();
	char[] c2 = s2.toCharArray();
	int length = s1.length();

	boolean[][][] is = new boolean[length][length][length];
	for (int i = 0; i < s1.length(); i++) {
		for (int j = 0; j < s2.length(); j++) {
			is[0][i][j] = (c1[i] == c2[j]);
		}
	}
	for (int k = 1; k < length; k++) {
		for (int m = length - k - 1; m >= 0; m--) {
			for (int n = length - k - 1; n >= 0; n--) {
				boolean r = false;
				for (int j = 1; j <= k && !r; j++) {
					if ((is[j - 1][m][n] && is[k - j][j + m][j + n])
							|| (is[j - 1][m][n + k + 1 - j] && is[k - j][m
									+ j][n])) {
						r = true;
					}
				}
				is[k][m][n] = r;
			}
		}
	}
	return is[length - 1][0][0];
}