【LeetCode练习题】Scramble String

时间:2022-05-29 11:39:44

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.

解题思路:

要满足isScramble(string s1,string s2),则必然满足isScramble(s11,s21)&&isScramble(s12,s22)或者isScramble(s11,s22)&&isScramble(s12,s21)

递归结束条件,当s1.compare(s2) == 0 时return false,即s1 和 s2 都只有一个字符且相等的时候。

当排序的sort1 ,sort2 不相等,即说明s1 和 s2 中的字符不同,return false,加上这个检查就可以大大的减少递归次数。否则就会超时。

每一次调用的s1 和 s2 的长度都是相等的,所以isScramble(s11,s21)&&isScramble(s12,s22)的时候s11.size() == s21.size(),

isScramble(s11,s22)&&isScramble(s12,s21)的时候s11.size() == s22.size()。

还有动态规划的解法,目前还不太熟,正在研究中……

代码如下:

class Solution {
public:
bool isScramble(string s1, string s2) {
string sort1 = s1,sort2 = s2;
sort(sort1.begin(),sort1.end());
sort(sort2.begin(),sort2.end());
if(sort1.compare(sort2) != )
return false;
if(s1.compare(s2) == )
return true; int len = s1.size();
for(int i = ; i < len; i++){
string s11 = s1.substr(,i);
string s12 = s1.substr(i);
string s21 = s2.substr(,i);
string s22 = s2.substr(i); if(isScramble(s11,s21)&&isScramble(s12,s22))
return true;
s21 = s2.substr(,len-i);
s22 = s2.substr(len-i);
if(isScramble(s11,s22)&&isScramble(s12,s21))
return true;
}
return false;
}
};