31. Next Permutation

 

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

The replacement must be in-place, do not allocate extra memory.

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

 

 

next_permutation的算法步骤（二找，一交换，一翻转）：

（1）    找到排列中最右边一个升序的首位位置i，x=nums[i]

（2）    找到排列中第i位右边最后一个比x大的位置j，y=nums[j]

（3）    交换x，y

（4）    把第i+1位到最后的部分进行翻转

1. class Solution {
   private:
       void reverse(vector<int>& nums,int from,int to){
           while(from<to){
               int temp=nums[from];
               nums[from++]=nums[to];
               nums[to--]=temp;
           }
       }
   public:
       void nextPermutation(vector<int>& nums) {
           int size=nums.size();
           int i=size-2;
           while(i>-1&&nums[i]>=nums[i+1])
               i--;
           if(i<0){
               reverse(nums,0,size-1);
               return;
           }
           int k=size-1;
           while(k>i&&nums[k]<=nums[i])
               k--;
           swap(nums[i],nums[k]);
           reverse(nums,i+1,size-1);
       }
   };