16. 3Sum Closest

16. 3Sum Closest     

     

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).

  问题描述：给定一个有n个数的数组，找出数组中3个数的和与给定值最接近的值。返回这3个数的和。

问题分析：先将前3个数的和赋值给result，利用3层循环遍历数组，最外层循环表示第一个数，中间层循环表示第二个数，最内层循环表示第3个数。因为这3个数代表数组中不同位置的数，所以循环时从当前位置的下一个开始循环。若当前循环中三个数的和sum比result更接近target（即Math.abs(sum-target) < Math.abs(result-target)），则sum为当前三个数的和最接近target的值，复制给result。

但是出现了测试用例都通过了但Time Limit Exceeded情况：

添加了语句： if(sum == target)  return sum;  就通过了

class Solution {    public int threeSumClosest(int[] nums, int target) {        int result = nums[0] + nums[1] + nums[2] ;        int sum = 0;        int sum1 = 0;        for(int i = 0;i<nums.length;i++){            for(int j = i+1;j<nums.length;j++){                sum = nums[i];                              sum += nums[j];                sum1 = sum;                for(int k = j+1;k<nums.length;k++) {                      sum = sum1;                                      sum += nums[k];                    if(sum == target)                        return sum;                    else if(Math.abs(sum-target) < Math.abs(result-target))                        result = sum;                                                     }                        }        }        return result;    }}