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.
Example:
Given array nums = [-1, 2, 1, -4], and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
找到三个数的和,使这个和距离target距离最近。
思路和3sum差不多,只不过需要记录一下最短距离即可。
还是先将数组排序,然后利用两个指针去遍历数组。多了一步记录距离的过程,注意使用abs()
class Solution:
def threeSumClosest(self, nums: List[int], target: int) -> int:
nums.sort()
closest=float('inf')
for i in range(len(nums)):
if i!=0 and nums[i]==nums[i-1]:continue
l, r=i+1,len(nums)-1
while l<r:
sumx=nums[i]+nums[l]+nums[r]
if sumx == target:
return target
if abs(sumx-target)<abs(closest-target):
closest=sumx
if sumx<target:
l+=1
else:
r-=1
return closest

本文介绍了一个经典的算法问题:在给定的整数数组中寻找三个数,使其和最接近给定的目标值。文章详细阐述了解决该问题的策略,即先对数组进行排序,然后使用双指针法遍历数组,通过比较当前和与目标值的距离来更新最接近的和。
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