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.
Example:
Input: nums = [-2,0,1,3], and target = 2
Output: 2
Explanation: 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?
class Solution { public int threeSumSmaller(int[] nums, int target) { if (nums == null || nums.length == 0) { return 0; } Arrays.sort(nums); int res = 0; for (int i = 0; i < nums.length - 2; i++) { int start = i + 1, end = nums.length - 1; while (start < end) { if (nums[i] + nums[start] + nums[end] < target) { // assume the end - 1... met res += end - start; start += 1; } else { end -= 1; } } } return res; } }
原文：https://www.cnblogs.com/xuanlu/p/12195893.html