No.016 3Sum Closest
No.016 3Sum Closest
mukekeheart
发布于 2018-02-27 15:33:04
发布于 2018-02-27 15:33:04
文章被收录于专栏：mukekeheart的iOS之旅
16. 3Sum Closest
- Total Accepted: 86565
- Total Submissions: 291260
- Difficulty: Medium
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).思路：
本题的解法和上一次3Sum的解法有点类似，我们可以先将数组排序，然后将数组中的数从左到右依次确定为第一个数，在其右边的数中寻找最接近的target的数即可。重要的是判断逻辑的思路。代码如下：
1 public int threeSumClosest(int[] nums, int target) {
2 if (nums == null || nums.length < 3) {
3 return 0;
4 }
5
6 Arrays.sort(nums);
7
8 int closest = nums[0]+nums[1]+nums[2] ;
9
10 for(int i = 0 ; i < nums.length-2 ; i++){
11 int l = i+1 ;
12 int r = nums.length-1 ;
13 while(l < r){
14 int sum = nums[i] + nums[l] + nums[r] ;
15 if(sum == target){
16 return sum ;
17 }else if(sum < target){
18 /*
19 * 三种情况：
20 * 1、closest < sum < target --->closest = sum
21 * 2、sum < target < closest --->| target-sum < closest-target ---> closest = sum
22 * | target-sum >= closest-target --> closest不变
23 * 3、sum <= closest <= target ---> closest不变
24 */
25 if(sum > closest){
26 //情况1
27 closest = sum ;
28 }else if(closest > target){
29 //情况2
30 if(target-sum < closest-target){
31 //情况2.1，
32 closest = sum ;
33 }
34 //情况2.2不用变化
35 }
36 //情况3不同变化
37 l++ ;
38 }else{
39 //分析同上
40 if(sum < closest){
41 closest = sum ;
42 }else if(closest < target){
43 if(sum-target <= target-closest){
44 closest = sum ;
45 }
46 }
47 r-- ;
48 }
49 }
50
51 }
52 return closest ;
53 }本文参与 腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。
原始发表：2016-07-29 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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