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? Show Company Tags Show Tags Show Similar Problems
小心这里是index triplets, 不是nums[i]数组元素的triplets, 所以3Sum那道题里面的跳过条件不用了
因为不关心每个index具体是什么,只关心个数,所以可以排序
public class Solution { public int threeSumSmaller(int[] nums, int target) { int res = 0; Arrays.sort(nums); for (int i=nums.length-1; i>=2; i--) { //if (i!=nums.length-1 && (nums[i]==nums[i+1])) continue; res += twoSum(nums, 0, i-1, target-nums[i]); } return res; } public int twoSum(int[] nums, int l, int r, int target) { int sum = 0; while (l < r) { if (nums[l]+nums[r] < target) { sum += r-l; l++; } else { r--; } } return sum; } }