题面
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.
给定数组,找出并返回最接近target的三个元素的和。可以假设,只有一个解。
样例
Given array nums = [-1, 2, 1, -4], and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
思路
我们在15题 3Sum 中做过,三数加和为target的问题,采用了Two-Point逼近的方法。本题我们稍加改动就可以解决。
1. 数组排序,固定一个元素i,通过两点法去[i+1, size()-1]中搜索另外两个数字,先计算他们的和;
2. 若sum == target,直接返回;若不等,就需要判断sum与 我们给的初值res 那个更加接近target,即判断 abs(sum - target) 与 abs(res - target)的大小,对res进行更新,另外注意 l 和 r 的更新。
3. 返回res.
源码
1 class Solution { 2 public: 3 int threeSumClosest(vector<int>& nums, int target) { 4 int len = nums.size(); 5 if(len < 3) 6 return 0; 7 //数组升序排序 8 sort(nums.begin(), nums.end()); 9 int res = nums[0]+nums[1]+nums[2]; 10 for(int i=0; i<len; i++) 11 { 12 if(i>0 && nums[i]==nums[i-1]) 13 i++;//避免i的重复,不必要 14 int l = i+1, r = len-1; 15 while(l < r)//两点法搜搜 16 { 17 int sum = nums[i] + nums[l] + nums[r]; 18 if(sum == target) 19 return sum; 20 else if(sum > target) 21 { 22 if(abs(sum - target) < abs(res - target)) 23 res = sum; 24 r--; 25 } 26 else 27 { 28 if(abs(sum - target) < abs(res - target)) 29 res = sum; 30 l++; 31 } 32 } 33 } 34 return res; 35 } 36 };