1.题目 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).2.解法 思路：从数组起始处遍历，下标为i，选取两个下标j,k作为两个指针遍历i+1到nums.length-1处的数组元素。比较temp=nums[i]+nums[j]+nums[k]以及初始值closest与target之间哪个距离更小，以此更新closest。外层循环每一次遍历之后是j++还是k–取决于temp大于target还是小于target。
时间复杂度：O（N^2）
public class Solution { public int threeSumClosest(int[] nums, int target) { int size = nums.length; if(size < 3){ return 0; } Arrays.sort(nums); int closest = nums[0] + nums[1] + nums[2]; for(int i = 0; i < size - 1; i++){ int j = i + 1; int k = size - 1; while(j < k){ int temp = nums[i] + nums[j] + nums[k]; if(temp == target){ return temp; }else if(temp > target){ --k; }else if(temp < target){ ++j; } if(Math.abs(target - temp) < Math.abs(target - closest)){ closest = temp; } } } return closest; } }