LeetCode 16.3Sum Closest

Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Example:

Given array nums = [-1, 2, 1, -4], and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

 1 class Solution4 {
 2 //    时间复杂度O(n3)
 3 //    public int threeSumClosest(int[] nums, int target) {
 4 //        Arrays.sort(nums);
 5 //        int closest = 65535, result = -65535;
 6 //        for (int i = 0; i < nums.length - 2; i++) {
 7 //            for (int low = i+1; low < nums.length - 1; low++) {
 8 //                for (int high = nums.length - 1; high > low; high--) {
 9 //                    int sum = nums[i] + nums[low] + nums[high];
10 //                    if (closest > Math.abs(target - sum)){
11 //                        closest = Math.abs(target - sum);
12 //                        result = sum;
13 //                    }
14 //                }
15 //            }
16 //        }
17 //        return result;
18 //    }
19 
20 //    时间复杂度O(n3) 前进了一小步
21 //    public int threeSumClosest(int[] nums, int target) {
22 //        Arrays.sort(nums);
23 //        int closest = 65535, result = -65535;
24 //        for (int i = 1; i < nums.length - 1; i++) {
25 //            for (int low = i-1; low >= 0; low--) {
26 //                for (int high = i+1; high < nums.length; high++) {
27 //                    int sum = nums[low] + nums[i] + nums[high];
28 //                    if (closest > Math.abs(target - sum)){
29 //                        closest = Math.abs(target - sum);
30 //                        result = sum;
31 //                    }
32 //                }
33 //            }
34 //        }
35 //        return result;
36 //    }
37 
38 //  前进了一大步
39     /*
40     可以看出while与for的区别：
41         while更灵活，能够根据情况控制step；for则不行
42      */
43     public int threeSumClosest(int[] nums, int target) {
44         Arrays.sort(nums);
45         int closest = 65535, result = -65535;
46         for (int i = 0; i < nums.length - 2; i++) {
47             int low = i + 1, high = nums.length - 1;
48             while (low < high){
49                 int sum = nums[i] + nums[low] + nums[high];
50                 if (closest >Math.abs(target-sum)){
51                     closest = Math.abs(target-sum);
52                     result = sum;
53                 }
54                 if (sum > target){
55                     --high;
56                 }else if(sum < target){
57                     ++low;
58                 }else {//正好相等
59                     return sum;
60                 }
61             }
62         }
63         return result;
64     }
65 }