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).
解题思路：
这道题其实跟leetcode15那道题的思路基本是一样的，就是那道题是找相等的，这道题是找相差最小的。
所以我的思路跟上道题差不多，先进行排序，然后拿出第一个数，再剩下的树中找到两个数，跟第一个数的和与目标值最相近的。不断递归即可。
当找到三个数的和等于目标值时，直接返回即可。
有一个tip就是，后面两个数，设两个下标，一个指向较小的数（j：初始值为i+1），一个指向较大的数（k：初始值为length-1），当三个数的和大于目标值时，k = k-1；当三个数的和小于目标值时，j = j+1。
[java] view plain copy print ? public class Solution { public int threeSumClosest(int[] nums, int target) { int result = 0; int dif = 2147483647; Arrays.sort(nums); for (int i = 0;i<nums.length-2;i++) { int j = i+1; int k = nums.length-1; while(j != k) { int sum = nums[j] + nums[k] + nums[i]; int diff = sum - target; if(Math.abs(diff)<dif) { result = sum; dif = Math.abs(diff); } if(Math.abs(diff) == 0) return result; if(diff>0) k--; else j++; } } return result; } } public class Solution { public int threeSumClosest(int[] nums, int target) { int result = 0; int dif = 2147483647; Arrays.sort(nums); for (int i = 0;i<nums.length-2;i++) { int j = i+1; int k = nums.length-1; while(j != k) { int sum = nums[j] + nums[k] + nums[i]; int diff = sum - target; if(Math.abs(diff)<dif) { result = sum; dif = Math.abs(diff); } if(Math.abs(diff) == 0) return result; if(diff>0) k--; else j++; } } return result; } }
