31. Next Permutation

Problem statement:

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:

• loop from the back of the array.
• record the maximum of all traversec elements.
• if current element is less than the maximum value
• find the element which is just greater than it from it`s back.
• swap these two elements and breaks
• if we did not do any swap, that means current element is already the maximum value for current nums. the return value is the reverse result

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        if(nums.size() <= 1){
            return;
        }
        int size = nums.size();
        int max_val = nums[nums.size() - 1];
        bool find = false;
        for(int ix = nums.size() - 2; ix >= 0; ix--){
            if(max_val > nums[ix]){
                sort(nums.begin() + ix + 1, nums.end());
                for(int i = ix + 1; i < size; i++){
                    // find the element just greater than nums[ix]
                    if(nums[i] > nums[ix]){
                        // swap these two elements
                        swap(nums[i], nums[ix]);
                        find = true;
                        // find the final solution and exit
                        break;
                    }
                }
                break;
            } else {
                max_val = nums[ix];
            }
        }
        // if current arrange is already the maximum value
        // reverse it to the mimimum value
        if(!find){
            reverse(nums.begin(), nums.end());
        }
        return;
    }
};