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 and use only constant 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

It really helps to draw this one out on grid. I was able to figure out the swapping logging, but missed the reversing part in my analysis.

Found this pretty concise solution here

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void nextPermutation(vector<int>& nums) {
    auto i = is_sorted_until(nums.rbegin(), nums.rend());
    if (i != nums.rend())
        swap(*i, *upper_bound(nums.rbegin(), i, *i));
    reverse(nums.rbegin(), i);
}

Other Good solutions