LeetCode:Next Permutation

problem:

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

solution

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        
        //从后往前扫描找到第一个不是升序的位置
        int par_pos=-1;
        for(int i=nums.size()-1;i>=0;i--)
        {
            if(nums[i]>nums[i-1])
                {   
                    par_pos=i-1;
                    break;
                }
            
        }
        //如若没找到par_pos 则已经为最后一个 只需要直接反转即可
        if(par_pos<0)  
        {
           reverse(nums.begin(),nums.end());
           return;
        }
        
        int cha_pos=-1;
        //从后往前找到第一个比par_pos位置的元素大的位置 cha_pos  
        for(int i=nums.size()-1;i>par_pos;i--)
       {
           if(nums[i]>nums[par_pos])
           {
             cha_pos=i;
             break;   
           }
       }
      //交换par_pos和cha_pos ,反转par_pos之后的元素
       swap(nums[par_pos],nums[cha_pos]);
       reverse(nums.begin()+par_pos+1,nums.end());        
       
    }
};