LeetCode 3Sum Smaller

原题链接在这里：https://leetcode.com/problems/3sum-smaller/

题目：

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]

题解：

与3Sum类似. 计算和小于target的组合个数，重复的也要算. 若是nums[i] + nums[j] + nums[k] 符合要求，那么count += k-j, j++. 举例{-2, 0, 1, 3}. target = 4. {-2, 0, 3}符合要求, {-2, 0, 1}也符合要求，一次加上两个.

若是不符合要求说明等于或者大于target了, k--.

Time Complexity: O(n^2). Space: O(1).

AC Java:

 1 public class Solution {
 2     public int threeSumSmaller(int[] nums, int target) {
 3         if(nums == null || nums.length == 0){
 4             return 0;
 5         }
 6         Arrays.sort(nums);
 7         int count = 0;
 8         for(int i = 0; i<nums.length-2; i++){
 9             int j = i+1;
10             int k = nums.length-1;
11             while(j<k){
12                 if(nums[i] + nums[j] + nums[k] >= target){
13                     k--;
14                 }else{
15                    count += k-j;
16                    j++;
17                 }
18             }
19         }
20         return count;
21     }
22 }