C ++ algorithms: ---- arranged next to find the law

topic:

Achieve a function of the acquired arrangement, a given algorithm requires rearranging the sequence number next in lexicographically larger arrangement.
If the next larger arrangement does not exist, the smallest number of rearranged alignment (i.e., ascending order).
You must place editing, allowing only constant additional space.

Example:

The following are some examples, the input column on the left, on the right column the corresponding output.
→ 1, 3,2, 2, 3
3,2,1 →, 2, 3
1,1,5 1,5,1 →

Code:

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        
        //思路：先考虑特殊情况，数组为从大到小排序的，例如：3->2->1，这种直接将数组反转变成从小到大
        //      当为1->4->6->5->3->2时，将从右到左第一个a[i] > a[i-1]的a[i-1]替换成右边略大于他的数a[j]，然后将a[j]右边的数从小到大排序，例如：将4替换成5，然后将3->2从小到大排序
        int i, j, temp;
        
        //找到a[i-1]
        for(i = nums.size()-1; i > 0; i--)
            if(nums[i-1] < nums[i])
                break;

        if(i != 0)
        {
             //找到a[j]
            for(j = nums.size()-1; j > i; j--)
                if(nums[j] > nums[i-1])
                    break;
            
            //交换a[i-1]和a[j]
            temp = nums[i-1];
            nums[i-1] = nums[j];
            nums[j] = temp;
        }
        else
        {
            j = -1;
        }
        
        //从大到小排序a[j]后面的元素
        for(j = i; j <= (nums.size()-1+i)/2; j++)
        {
            temp = nums[j];
            nums[j] = nums[nums.size()-1-(j-i)];
            nums[nums.size()-1-(j-i)] = temp;
        }            
        
    }
};