No.31. Next Permutation

//Code
class Solution {
public:
    void nextPermutation(vector<int>& nums) 
    {
        if(nums.size()==0||nums.size()==1)
            return ;
        next(nums,1);
        return ;
    }
    
    
    void next(vector<int> & nums, int n)  // find next arage
    {
        int len=nums.size();
        if(n==len)                         //to the head, 
        {
            rev(nums);
            return ;
        }
        else if(nums[len-n-1]>=nums[len-n])         //
        {
            n++;
            next(nums,n);
        }
        else
        {    int i;
          for(i=len-1;i>=len-n;i--)              //find the smallest in last n nums;
          {
              if(nums[i]>nums[len-n-1])
                  break;
          }
            int x=nums[len-n-1];                   //swap
            nums[len-n-1]=nums[i];
            nums[i]=x;
            sort(nums.begin()+len-n,nums.end());     rearrage
            return ;
         
        }
    }
    
    void rev(vector<int> & nums)                  //no,reverse
    {
        sort(nums.begin(),nums.end());
        return ;
    }
};