3Sum Smaller leetcode

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]

用3Sum的方法 a=num[i] + num[l] + num[r]
当a < target时候 left++, 但是注意此时对于num[i] + num[l] + num[l + 1 ---> r] 都满足<target
所以这个时候指针右移 count+ right - left;

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