【LeetCode】87. Scramble String

时间:2021-01-28 13:28:16

题目:

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.

提示:

这道题可以用递归方法较为简单地解决,递归的形式可以这样:

// recursive solution, the idea is based on the binary tree struct.
for (int i = ; i < s1.length(); ++i) {
if (isScramble(s1.substr(, i), s2.substr(, i)) && isScramble(s1.substr(i), s2.substr(i))) {
return true;
}
if (isScramble(s1.substr(, i), s2.substr(s2.length() - i)) && isScramble(s1.substr(i), s2.substr(, s2.length() - i))) {
return true;
}
}

结合题目的意思,如果两个单词是scramble的话,那么他们可能会有非叶子节点进行了逆序的操作,因此我们需要同时判断:

  • s1的左子节点与s2的左子节点,s1的右子节点与s2的右子节点
  • s1的左子节点与s2的右子节点,s1的右子节点与s2的左子节点

另外在进行递归前,需要对字符串是否相等,长度是否相等,是否是变位词进行判断。

代码:

class Solution {
public:
bool isScramble(string s1, string s2) {
// equal ?
if (s1 == s2) {
return true;
}
// if length are differrent, they can not be scramble
if (s1.length() != s2.length()) {
return false;
}
// just like anagram
vector<int> a(, );
for (int i = ; i < s1.length(); ++i) {
++a[s1[i]];
--a[s2[i]];
}
for (int i = ; i < s1.length(); ++i) {
if (a[s1[i]] != ) {
return false;
}
}
// recursive solution, the idea is based on the binary tree struct.
for (int i = ; i < s1.length(); ++i) {
if (isScramble(s1.substr(, i), s2.substr(, i)) && isScramble(s1.substr(i), s2.substr(i))) {
return true;
}
if (isScramble(s1.substr(, i), s2.substr(s2.length() - i)) && isScramble(s1.substr(i), s2.substr(, s2.length() - i))) {
return true;
}
}
return false;
}
};