[LeetCode] 3Sum Closest 最近的三数之和 Python

3Sum Closest：

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

简单来说，寻找三数之和最接近target的答案，例子如下：

 For example, given array S = {-1 2 1 -4}, and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

 解决方法：

和 上一篇3Sum的想法类似，因此代码也是在那个的基础上修改的。只不过因为需要求最近的，而不是固定的，因此所有的判定都需要修改为判断与与target做差后的绝对值， 因为代码大构架和3Sum类似，因此时间复杂度还是O(N^2)，代码如下：

class Solution(object):    def threeSumClosest(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: int        """        nums.sort()        first=[]        i=0        Max=0        while(i<len(nums)-2):            if(nums[i]!=nums[i-1] or i==0):                left=i+1                right=len(nums)-1                while(left<right):                    if(abs(nums[left]+nums[right]+nums[i]-target)==0):                        Max=target                        break                    if(i==0 and left==1 and right==len(nums)-1):                        Max=nums[left]+nums[right]+nums[i]                    if(abs(nums[left]+nums[right]+nums[i]-target)<abs(Max-target)):                        first.append([nums[i],nums[left],nums[right]])                        Max=nums[left]+nums[right]+nums[i]                        while(left<right and nums[left]+nums[right]+nums[i]<target):                            left+=1                            if(nums[left]!=nums[left-1]):                                break                        while(left>right and nums[left]+nums[right]+nums[i]>target):                            right-=1                            if(nums[right]!=nums[right+1]):                                break                    elif(nums[left]+nums[right]+nums[i]>target):                        right-=1                    elif(nums[left]+nums[right]+nums[i]<target):                        left+=1            i+=1        return Max