【LeetCode】#87扰乱字符串(Scramble String)

【LeetCode】#87扰乱字符串(Scramble String)

题目描述

给定一个字符串 s1，我们可以把它递归地分割成两个非空子字符串，从而将其表示为二叉树。
下图是字符串 s1 = “great” 的一种可能的表示形式。
great
/
gr eat
/ \ /
g r e at
/
a t
在扰乱这个字符串的过程中，我们可以挑选任何一个非叶节点，然后交换它的两个子节点。
例如，如果我们挑选非叶节点 “gr” ，交换它的两个子节点，将会产生扰乱字符串 “rgeat” 。
rgeat
/
rg eat
/ \ /
r g e at
/
a t
我们将 "rgeat” 称作 “great” 的一个扰乱字符串。
同样地，如果我们继续将其节点 “eat” 和 “at” 进行交换，将会产生另一个新的扰乱字符串 “rgtae” 。
rgtae
/
rg tae
/ \ /
r g ta e
/
t a
我们将 "rgtae” 称作 “great” 的一个扰乱字符串。
给出两个长度相等的字符串 s1 和 s2，判断 s2 是否是 s1 的扰乱字符串。

示例

示例 1:

输入: s1 = “great”, s2 = “rgeat”
输出: true

示例 2:

输入: s1 = “abcde”, s2 = “caebd”
输出: false

Description

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.

Example

Example 1:

Input: s1 = “great”, s2 = “rgeat”
Output: true

Example 2:

Input: s1 = “abcde”, s2 = “caebd”
Output: false

解法

class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1.length()!=s2.length())
            return false;
        if(s1.equals(s2))
            return true;
        char[] c1 = s1.toCharArray();
        char[] c2 = s2.toCharArray();
        Arrays.sort(c1);
        Arrays.sort(c2);
        if(!Arrays.equals(c1,c2))
            return false;
        for (int i = 1; i < s1.length(); i++) {
            if(isScramble(s1.substring(0,i),s2.substring(0,i))
                    && isScramble(s1.substring(i),s2.substring(i)))
                return true;
            if(isScramble(s1.substring(0,i),s2.substring(s2.length()-i))
                    && isScramble(s1.substring(i),s2.substring(0,s2.length()-i)))
                return true;
        }
        return false;
    }
}