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).
思路:和上一题差不多,将三个数的和减去target取绝对值,每次找出误差最小值及对应的三个数的和。
public class Solution { public int threeSumClosest(int[] nums, int target) { int len=nums.length; int closestVal=Integer.MAX_VALUE; int closestSum=nums[0]+nums[1]+nums[2]; Arrays.sort(nums); int currentNum=nums[0]; for(int i=0;i<len-2;i++){ if(i>0 && nums[i]==currentNum) continue; int begin=i+1,end=len-1; while(begin<end){ int sum=nums[i]+nums[begin]+nums[end]; int absVal=Math.abs(sum-target); if(absVal<closestVal){ closestVal=absVal; closestSum=sum; }else if(sum<target){ begin++; }else end--; } currentNum=nums[i]; } return closestSum; } }