https://leetcode-cn.com/problems/3sum-closest/
# -*- coding:utf-8 -*- class Solution(object): def threeSumClosest(self, nums, target): """ :type nums: List[int] :type target: int :rtype: int """ dis = float('inf') res = [] nums.sort() for i in range(len(nums) - 2): left = i + 1 right = len(nums) - 1 while left < right: if abs(nums[i] + nums[left] + nums[right] - target) < dis: dis = abs(nums[i] + nums[left] + nums[right] - target) res = [nums[i], nums[left], nums[right]] # 判断指针移动方向 if nums[i] + nums[left] + nums[right] - target < 0: left += 1 elif nums[i] + nums[left] + nums[right] - target > 0: right -= 1 else: return target return sum(res) if __name__ == '__main__': nums = [-1, 2, 1, -4] nums = [1, 2, 4, 8, 16, 32, 64, 128] target = 82 print(Solution().threeSumClosest(nums, target))
思路:这道题和上面的那个三数之和十分相似,一样是使用指针搜索,只不过不再以和为0作为判断,对应的引入一个变量dis用来表示当前集合和target的距离,同时根据当前集合的和 和 target的大小关系决定指针的移动。
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