Leetcode: Scramble String
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.
分析:这道题用递归的方法解逻辑清晰且不易出错,我们将原来的字符串s1分为左边长为i,右边长为l-i (l为字符串总长度)两部分,用递归的思想,如果下面两种情况一种成立则s2是s1的scramble string。
1)s1左端长为i的子串与s2左端长为i的子串互为scramble string, s1右端长为l-i的子串与s2右端长为l-i的子串互为scramble string。
2)s1左端长为i的子串与s2右端长为i的子串互为scramble string, s1右端长为l-i的子串与s2左端长为l-i的子串互为scramble string。
代码如下:
class Solution { public: bool isScramble(string s1, string s2) { if(s1 == s2) return true; //if characters in s1 and s2 are different, return false string str1 = s1, str2 = s2; sort(str1.begin(), str1.end()); sort(str2.begin(), str2.end()); if(str1 != str2) return false; int l = s1.length(); //iterate length of left-edian substring of s1 for(int i = 1; i <= l - 1; i++){ if(isScramble(s1.substr(0,i), s2.substr(l - i))&&isScramble(s1.substr(i), s2.substr(0, l-i)) || isScramble(s1.substr(0,i), s2.substr(0,i))&&isScramble(s1.substr(i), s2.substr(i))) return true; } return false; } };