[leetcode]87. Scramble String(Java)

https://leetcode.com/problems/scramble-string/#/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.

package go.jacob.day709; /** * 87. Scramble String * @author Jacob * */ public class Demo1 { /* * 参考自leetcode网友：@baifriend * 题意：只能交换非叶节点的两个子节点，不是任意交换叶节点！ */ public boolean isScramble(String s1, String s2) { if (s1.equals(s2)) return true; int[] letters = new int[26]; for (int i = 0; i < s1.length(); i++) { letters[s1.charAt(i) - 'a']++; letters[s2.charAt(i) - 'a']--; } for (int i = 0; i < 26; i++) if (letters[i] != 0) 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; } /* * 天真的解法：比如abc和bca */ public boolean isScramble_1(String s1, String s2) { if (s1.equals(s2)) return true; int[] letters = new int[26]; for (int i = 0; i < s1.length(); i++) { letters[s1.charAt(i) - 'a']++; letters[s2.charAt(i) - 'a']--; } for (int i = 0; i < 26; i++) { if (letters[i] != 0) return false; } int count = 0; for (int i = 0; i < s1.length(); i++) { if (s1.charAt(i) != s2.charAt(i)) count++; } return count % 2 == 0; } }