蘇州大學大三考研狗,保研也不穩,水橋杯還要國賽,日常更新博客,不出意外每天更新一題LeetCode hard難度
題意:給一二叉樹,對於非葉子節點可以交換左右孩子結點,從而形成了新的樹,繼而每個結點代表的字符串打亂順序,問你給你兩個字符串s1與s2,問s2是否可以由s1做上述操作得到。
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"
.
思路:首先,既然是二叉樹,那麼一分爲二的過程中,左結點與右結點應該是擁有自此之後不變的字符種類與數量的,我們需要做的是找到這樣的分界點,然後判斷當前深度左右結點是否滿足種類與個數不變的原則。至於遍歷深度,用遞推即可,這裏需要注意的是要剪枝,不然遞推層數過多會tle。
AC code:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
class Solution
{
public:
bool isScramble(string s1, string s2)
{
//特判
if(s1.size()!=s2.size())
return false;
if(s1==s2)
return true;
vector<int>cnt(30,0);
for(auto i:s1)
cnt[i-'a']++;
for(auto i:s2)
cnt[i-'a']--;
//剪枝
for(int i=0;i<26;i++)
if(cnt[i]!=0)
return false;
//遞歸
for(int i=1;i<s1.size();i++)
if( (isScramble(s1.substr(0,i),s2.substr(0,i)) && isScramble(s1.substr(i,s1.size()-i),s2.substr(i,s1.size()-i)))
|| (isScramble(s1.substr(0,i),s2.substr(s1.size()-i)) && isScramble(s1.substr(i),s2.substr(0,s1.size()-i))) )
return true;
return false;
}
};
int main()
{
string s1,s2;
cin>>s1>>s2;
Solution tmp;
cout<<tmp.isScramble(s1,s2)<<endl;
return 0;
}