一、题目说明
题目是31. Next Permutation,英文太差看不懂,翻译了一下。才知道是求字典顺序下的下一个排列,不允许使用额外空间。题目难度是Medium!
二、我的实现
首先要进一步理解题目,以1->2->3
为例,字典顺序如下:
(1) 1->2->3;
(2) 1->3->2;
(3) 2->1->3;
(4) 2->3->1;
(5) 3->1->2;
(6) 3->2->1;
(7) 1->2->3;
如何从(1)-> (2) ->(3)-> (4) ->(5)-> (6) ->(7)实现状态转换?以(3)->(4)为例:
从列表lists的最右边起,
if(lists[t] < lists[t-1]) {
swap(lists[t-1],max{lists[t]...lists[listSize-1]})
sort(lists[t],lists[listSize-1]);
}
从(6)->(7),sort(lists[0],lists[listSize-1])即可。
代码如下:
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
class Solution {
public:
void nextPermutation(vector<int>& nums){
if(nums.size()<=1) return ;
bool flag = false;
for(int t=nums.size()-1;t>0;t--){
if(nums[t]>nums[t-1]){
//find the smallest between nums[t] to nums[t-1]
flag = true;
int max = nums[t];
int maxIndex = t;
for(int k=nums.size()-1;k>=t;k--){
if(nums[t-1]<nums[k]){
max = nums[k];
maxIndex = k;
break;
}
}
int tmp = nums[t-1];
nums[t-1] = nums[maxIndex];
nums[maxIndex] = tmp;
//从t..size()-1重新排序
int len = nums.size()-t;
for(int s=0;s<(len+1)/2;s++){
tmp = nums[t+s];
nums[t+s] = nums[nums.size()-s-1];
nums[nums.size()-s-1] = tmp;
}
break;
}
}
if(!flag){
int tmp,len = nums.size();
for(int t=0;t<(len+1)/2;t++){
tmp = nums[t];
nums[t] = nums[len-t-1];
nums[len-t-1] = tmp;
}
}
}
};
int main(){
Solution s;
vector<int> v;
v = {1,3,2};
s.nextPermutation(v);
for(vector<int>::iterator it=v.begin();it!=v.end();it++){
cout<<*it<<" ";
}
cout<<endl;
v = {5,4,7,5,3,2};
s.nextPermutation(v);
for(vector<int>::iterator it=v.begin();it!=v.end();it++){
cout<<*it<<" ";
}
cout<<endl;
v = {3,2,1};
s.nextPermutation(v);
for(vector<int>::iterator it=v.begin();it!=v.end();it++){
cout<<*it<<" ";
}
cout<<endl;
v = {1,5,1};
s.nextPermutation(v);
for(vector<int>::iterator it=v.begin();it!=v.end();it++){
cout<<*it<<" ";
}
cout<<endl;
return 0;
}
三、改进措施
提交后,性能如下:
Runtime: 8 ms, faster than 78.45% of C++ online submissions for Next Permutation.
Memory Usage: 8.6 MB, less than 88.17% of C++ online submissions for Next Permutation.
差不多了,就不优化了。