LeetCode:Scramble String

Scramble String

Total Accepted: 47175 Total Submissions: 176273 Difficulty: Hard

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.

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思路：

如：

s1 = "great",s2 = "rgtae"

  i 0 1 2 3 4

s1：g r e a t

s2：r g t a e

如当i=2时，比较以下四部分：

1）gr与rg

2）eat与tae

3）gr与ae

4）eat与rgt

java code:

public class Solution {    public boolean isScramble(String s1, String s2) {                if(s1.equals(s2)) return true;                int[] hash = new int[26];        for(int i=0;i<26;i++) hash[i]=0;                int len = s1.length();                for(int i=0;i<len;i++) {            hash[s1.charAt(i)-'a']++;            hash[s2.charAt(i)-'a']--;        }                for(int i=0;i<26;i++)            if(hash[i] != 0) return false;                            for(int i=1;i<len;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(len-i))                && isScramble(s1.substring(i),s2.substring(0,len-i)))                return true;        }        return false;    }}