1. 题目
给定一个包括 n 个整数的数组 nums 和 一个目标值 target。找出 nums 中的三个整数，使得它们的和与 target 最接近。返回这三个数的和。假定每组输入只存在唯一答案。
例如，给定数组 nums = [-1，2，1，-4], 和 target = 1.与 target 最接近的三个数的和为 2. (-1 + 2 + 1 = 2).
来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/3sum-closest
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
2. 解题
- 先排序
- 固定左端的数nums[i],0<=i<=n−3nums[i], 0<= i <= n-3nums[i],0<=i<=n−3
- 然后让start=nums[i+1],end=nums[n−1]start = nums[i+1], end = nums[n-1]start=nums[i+1],end=nums[n−1]，比较与target的差距，区间向中间收缩
- 循环，移动左端的 i
class Solution { //C++
public:int threeSumClosest(vector<int>& nums, int target) {sort(nums.begin(), nums.end());int i, s, e, minDiffer = INT_MAX, record;int cursum = nums[0]+nums[1]+nums[2];if(target <= cursum)return cursum;for(int i = 0; i < nums.size()-2; ++i){s = i+1; //右边区间起点e = nums.size()-1; //右边区间终点while(s < e){cursum = nums[i]+nums[s]+nums[e];if(abs(cursum-target) < minDiffer){minDiffer = abs(cursum-target);record = cursum;}if(cursum < target)++s; //收缩区间else if(cursum == target)return cursum;else--e; //收缩区间}}return record;}
};
class Solution:# py3def threeSumClosest(self, nums: List[int], target: int) -> int:nums.sort()minDiffer, record = float('inf'), 0cursum = nums[0]+nums[1]+nums[2];if target <= cursum:return cursumn = len(nums)cursum = nums[n-3]+nums[n-2]+nums[n-1]if target >= cursum:return cursumfor i in range(n-2):s = i+1e = n-1while s < e:cursum = nums[i]+nums[s]+nums[e]if abs(cursum-target) < minDiffer:minDiffer = abs(cursum-target)record = cursumif cursum < target:s += 1elif cursum == target:return cursumelse:e -= 1return record
)
