3Sum Smaller

Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.

For example, given nums = [-2, 0, 1, 3], and target = 2.

Return 2. Because there are two triplets which sums are less than 2:

[-2, 0, 1]
[-2, 0, 3]

Follow up:
Could you solve it in O(n²) runtime?

//Sort it , and then use two pointers algorithm
class Solution {
public:
    int threeSumSmaller(vector<int>& nums, int target) {
        if(nums.size() < 3) return 0;
        sort(nums.begin(), nums.end());
        int count = 0;
        for(int i = 0; i < nums.size() - 2; i++){
            int j = i + 1;
            int k = nums.size() - 1;
            while(j < k){ 
                int sum = nums[i] + nums[j] + nums[k];
                if(sum < target){
                    count += k - j;
                    j++;
                }
                else{
                    k--;
                }
            }
        }
        return count;
    }
};