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?
[分析]
延续3Sum & 3Sum Closest的基本套路,先排序输入数组,固定第一个数后在剩余数组区间中寻找所有满足条件的另外两个数,使用two pointers技巧进行寻找。
关键点是一旦找到 i, j 满足nums[i] + nums[j] + nums[k] < target, 因为数组已排序,则(i, j-1, k),(i, j - 2, k)...(i, i+1, k)均是满足条件的解,这些解总共是 j - i个,而无需一个一个check它们。
此外需注意此题不需要去重处理,一开始画蛇添足进行了去重处理……
很开心思路和leetcode讨论版的最佳思路一致~
public class Solution { public int threeSumSmaller(int[] nums, int target) { if (nums == null || nums.length < 3) return 0; Arrays.sort(nums); int counter = 0; int end = nums.length - 3; for (int k = 0; k <= end; k++) { int i = k + 1, j = nums.length - 1; while (i < j) { int sum3 = nums[i] + nums[j] + nums[k]; if (sum3 < target) { counter += j - i; i++; } else { while (i < --j && nums[j] == nums[j + 1]); } } } return counter; } }