[LeetCode] 87. Scramble String Java

Posted 2020-10-17 BrookLearnData

题目：

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.

题意及分析：一个字符串可以递归分割成任意两个非空字符串，任意交换其中两个非叶子节点，得到一个新字符串，这个新字符串就称作字符串的scrambled strin。给出两个字符串S1和s2，判断s2是不是s1的scramble string，如果s2是s1的scramble串，那么必然存在一个在s1上的长度l1，将s1分成s11和s12两段，同样有s21和s22。那么只有两种可能：（1）s11和s21是scramble且s12和s22是scramble （2）s11和s22是scramble且s12和s21是scramble

代码：

class Solution {
    public boolean isScramble(String s1, String s2) {
        if (s1.equals(s2)) return true;

        int length = s1.length();
        int[] count = new int[26];
        for(int i=0;i<length;i++){
            count[s1.charAt(i)-‘a‘] ++;
            count[s2.charAt(i)-‘a‘] --;
        }
        for(int i=0; i<26; i++) {
            if(count[i]!=0)
                return false;
        }

        for(int i=1;i<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(length-i))
                    && isScramble(s1.substring(i), s2.substring(0,length-i))) return true;
        }
        return false;
    }
}