leetcode-87. Scramble String

leetcode-87. Scramble String

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.

做这题的时候，总想着二分去解实际上不应该这样。虽然似乎也能做出来，但是会出现栈溢出。至少这样消耗的资源是很多的。

这里的基本方法就是累积字符数量，
然后对不同的位置做切分去找有没有满足条件的，如果有则直接返回。
从方法上来说其实综合了动态规划，hash表和DFT的方法是很好的综合性题目

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

        int[] list = new int[26];
        for(int i = 0 ; i < s1.length() ;i++){
            list[s1.charAt(i)-'a']++;
            list[s2.charAt(i)-'a']--;
        }

        for(int i : list)
            if(i!=0) return false;

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