31. Next Permutation

本文摘自 LeetCode 题解

3 1. 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

关于next_permutation的解法，比较好的说明是下图

代码：

class Solution {
public:
        void nextPermutation(vector<int>& nums) {
            next_permutation(nums.begin(), nums.end());
        }

    private:
        template <typename BidiIt>
        bool next_permutation(BidiIt first, BidiIt last) {
            //get a reversed range to simplify reversed traversal
            const auto rfirst = reverse_iterator(last);
            const auto rlast = reverse_iterator(first);

            //Begin from the second last element to the first element
            auto pivot = next(rfirst);

            //Find `pivot`, which is the first element that is no less than its successor. `Pre` is used since `pivort` is a `reversed_iterator`.
            while(pivot != rlast && *pivot >= *prev(pivot)) 
                ++pivot;


            //NO such element found, current sequence is already the largets
            //permutation. the rearrange to the first permutation and return false.
            if(pivot == rlast) {
                reverse(rfirst, rlast);
                return false;
            }


            //Scan from right to left, find the first element that is greater than `pivot`.
            auto change = find_if(rfirst, pivot, bind1st(less <int> (), *pivot));

            swap(*change, *pivot);
            reverse(rfirst, pivot);

            return true;
        }

};