标签：present lse input eterm and miss bool code sub
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.
Example 1:
Input: s1 = "great", s2 = "rgeat" Output: true
Example 2:
Input: s1 = "abcde", s2 = "caebd" Output: false
AC code:
class Solution {
public:
bool isScramble(string s1, string s2) {
if (s1 == s2) return true;
int len = s1.length();
vector<int> v(26, 0);
for (int i = 0; i < len; ++i) {
v[s1[i]-‘a‘]++;
v[s2[i]-‘a‘]--;
}
for (int i = 0; i < 26; ++i) {
if (v[i] != 0) return false;
}
for (int i = 1; i < len; ++i) {
if (isScramble(s1.substr(0, i), s2.substr(0, i)) &&
isScramble(s1.substr(i), s2.substr(i))) return true;
if (isScramble(s1.substr(0, i), s2.substr(len-i)) &&
isScramble(s1.substr(i), s2.substr(0, len-i))) return true;
}
return false;
}
};
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Scramble String.
标签：present lse input eterm and miss bool code sub
原文地址：https://www.cnblogs.com/ruruozhenhao/p/9858820.html
