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 conditionnums[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?

这道题的解法与所有的3 sum，4 sum一样。唯一的区别在于两边往中间找的时候，如果小于dif，那么中间所有的都满足小于的情况，比如[3,1,0,-2], 4。 Sort完是[-2,0,1,3], 一开始选-2时，dif为6，left为1，right为3，小于dif，则以right为一个，left从左一直到right-1的组合都是满足要求的。

public int ThreeSumSmaller(int[] nums, int target) {        Array.Sort(nums);        var res =0;        int size = nums.Count();        for(int i =0;i< size;i++)        {            int num = nums[i];            int dif = target - num;            int left =i+1;            int right = size -1;            while(left<right)            {                if(nums[left]+nums[right] < dif)                {                    res+=right-left;                    left++;                }                else right--;            }        }        return res;    }

259. 3Sum Smaller