Leetcode: 87. Scramble String
2017-06-15 10:39
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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.
思路
递归使用一个hash表来判断两个字符串的字符是否相同,若不同,直接返回false
由于不知道左右子树从哪里划分,所以循环从1-len来划分左右子树,注意判断两个串的两种情况
代码
class Solution { public: bool isScramble(string s1, string s2) { if(s1 == s2) return true; return helper(s1, s2); } bool helper(string s1, string s2){ if(s1 == s2) return true; vector<int> count(26, 0); int len = s1.size(); for(int i = 0; i < len; ++i){ count[s1[i] - 'a']++; count[s2[i] - 'a']--; } for(int i = 0; i < 26; ++i) if(count[i]) return false; for(int i = 1; i < len; ++i){ if(helper(s1.substr(0, i), s2.substr(0, i)) && helper(s1.substr(i, len - i), s2.substr(i, len - i)) || helper(s1.substr(0, i), s2.substr(len - i, i)) && helper(s1.substr(i, len - i), s2.substr(0, len - i))) return true; } return false; } };
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