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?

Solution：two pointers

思路： Just basically same with other "sum"
但要求的是的count，所以count += right - left. 雙指針和滿足<target的情況下，right左減到底也都滿足。然后再升加left，繼續(xù)此過(guò)程。
Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

public class Solution {
    
    public int threeSumSmaller(int[] nums, int target) {
        int count = 0;
        Arrays.sort(nums);
        
        int len = nums.length;
    
        for(int i = 0; i < len - 2; i++) {
            int left = i+1, right = len - 1;
            while(left < right) {
                if(nums[i] + nums[left] + nums[right] < target) {
                    count += right - left;
                    left++;
                } else {
                    right--;
                }
            }
        }
        
        return count;
    }
}