【leetcode每日刷题】87. Scramble String

本文链接： https://blog.csdn.net/m0_38103546/article/details/102652296

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 1:

Input: s1 = "great", s2 = "rgeat"
Output: true

Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false

import java.util.Arrays;

class num87 {
    public boolean isScramble(String s1, String s2) {
        if(s1.length() != s2.length())  return false;
        if(s1.equals(s2))    return true;
        char[] arr1 = s1.toCharArray();
        char[] arr2 = s2.toCharArray();
        Arrays.sort(arr1);
        Arrays.sort(arr2);
        if(!String.valueOf(arr1).equals(String.valueOf(arr2)))    return false;
        for(int i=1; i<s1.length(); i++){
            String s11 = s1.substring(0, i);
            String s12 = s1.substring(i);
            String s21 = s2.substring(0, i);
            String s22 = s2.substring(i);
            if(isScramble(s11, s21) && isScramble(s12, s22)) return true;
            s21 = s2.substring(s1.length()-i);
            s22 = s2.substring(0, s1.length()-i);
            if(isScramble(s11, s21) && isScramble(s12, s22)) return true;
        }
        return false;
    }
    public static void main(String[] args) {
        String s1 = "abcde";
        String s2 = "caebd";
        num87 solution = new num87();
        System.out.println(solution.isScramble(s1, s2));
    }
}

使用递归的方法：先判断两个字符串的长度，然后判断是否相等，再判断排序后的字符数组是否相等。先切割两个字符串，分别截取相同的长度，递归判断s1左边的和s2左边的字符串，s1右边和s2右边的字符串是否Scramble；再判断s1左边的和s2右边的字符串是否Scramble; 直到遍历完整个长度。

注：java使用的时候遇到问题，将字符串转化为字符数组然后再转化为字符串比较值，需要使用函数String.valueof()，另外，字符串的比较不能直接用==，需要使用equal函数，使用 == 判断的是地址，不是字符串的值。java中的对象都是引用值。

【leetcode每日刷题】87. Scramble String

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