CODE 46: Scramble String

最新推荐文章于 2021-01-27 05:50:24 发布

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最新推荐文章于 2021-01-27 05:50:24 发布

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分类专栏： leetcode

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本文介绍了一种判断两个字符串是否可以通过乱序操作相互转换的算法。该算法通过构建二叉树来表示字符串，并通过递归地交换节点的孩子来实现字符串的乱序。文章详细解释了算法的具体实现过程，并提供了一个Java方法用于判断两个字符串是否为乱序版本。

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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];
	}