16. 3Sum Closest
16. 3Sum Closest
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).
解题思路：
由于需要找出三个这样的数，因此维护三个下标i，j，k，我们可以先对这个列表进行排序。第一个下标i从0移动到n-1，对于每一个下标i，设定j的初始值为i+1，k的初始值为n-1，j和k的移动方式遵循以下规则：若三个数的和大于目标值target，则j增加1，否则k减少1，直到j与k相等为止。由于列表已经是排好序的，因此此做法是正确的。对于每一组这样的i，j，k，可以找出与目标值target相差最小的和值，其时间复杂度为O(n^2)，比暴力解法的时间复杂度O(n^3)要低。
代码
class Solution:
def threeSumClosest(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
nums.sort()
closest_sum = nums[0]+nums[1]+nums[2]
distance = abs(closest_sum-target)
n = len(nums)
for i in range(0,len(nums)-2):
j = i+1;
k = n-1;
while(j<k):
if abs(nums[i]+nums[j]+nums[k]-target)<distance:
distance = abs(nums[i]+nums[j]+nums[k]-target)
closest_sum = nums[i]+nums[j]+nums[k]
if (nums[i]+nums[j]+nums[k]) < target:
j+=1
else:
k-=1
return closest_sum