next permutaion算法

算法描述：
Find largest index i such that array[i − 1] < array[i].

Find largest index j such that j ≥ i and array[j] > array[i − 1].

Swap array[j] and array[i − 1].

Reverse the suffix starting at array[i].

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public:
//    void nextPermutation(vector<int>& nums) {
//        //find the fisrt non-increacing number from the end
//        if(nums.size() < 2) return;
//        vector<int>::reverse_iterator it = nums.rbegin();
//        while((it+1) != nums.rend() && *(it+1) >= *it) it++;
//        if(it+1 == nums.rend()){
//            reverse(nums.begin(), nums.end());
//        }
//        else{
//            //find the min number greater than *(it + 1)
//            vector<int>::reverse_iterator itb = it+1;
//            vector<int>::reverse_iterator ite = nums.rbegin();
//            while(ite != itb && *ite <= *itb) ite++;
//            int temp = *ite;
//            *ite = *itb;
//            *itb = temp;
//            reverse(nums.rbegin(), itb);
//        }
//    }

    void nextPermutation(vector<int>& nums) {
        if(nums.size() < 2) return;
        int pos = nums.size();
        while(--pos > 0){
            if(nums[pos] > nums[pos - 1]){
                int pivot = pos -1;
                int back = nums.size();
                while(nums[--back] <= nums[pivot]){};
                swap(nums[pivot], nums[back]);
                break;
            }
        }
        reverse(nums.begin()+pos,nums.end());
    }
};