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.
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).
public class Solution { public int threeSumClosest(int[] nums, int target) { int length=nums.length,Min=Integer.MAX_VALUE; Arrays.sort(nums); for(int i=0;i<length-2;i++){ if(i>0 && nums[i]==nums[i-1])continue; int begin=i+1,end=length-1; while(begin<end){ int sum=nums[i]+nums[begin]+nums[end]; if(Math.abs(sum-target)<Math.abs(Min))Min=sum-target; if(sum==target)return target; else if(sum>target)end--; else begin++; } } return Min+target; } }
int abs(int tar){ return tar>0?tar:-tar; } void quickSort(int* nums,int first,int end){ int l=first,r=end; if(first>=end)return; int temp=nums[l]; while(l<r){ while(l<r && nums[r]>=temp)r--; if(l<r)nums[l]=nums[r]; while(l<r && nums[l]<=temp)l++; if(l<r)nums[r]=nums[l]; } nums[l]=temp; quickSort(nums,first,l-1); quickSort(nums,l+1,end); } int threeSumClosest(int* nums, int numsSize, int target) { int begin,end,i,sum,Min=INT_MAX; quickSort(nums,0,numsSize-1); for(i=0;i<numsSize-2;i++){ if(i>0 && nums[i]==nums[i-1])continue; begin=i+1;end=numsSize-1; while(begin<end){ sum=nums[i]+nums[begin]+nums[end]; if(abs(sum-target)<abs(Min))Min=sum-target; if(sum==target)return target; else if(sum>target)end--; else begin++; } } return Min+target; }
class Solution { public: int threeSumClosest(vector<int>& nums, int target) { int length=nums.size(),Min=2147483647; quickSort(nums,0,length-1); for(int i=0;i<length-2;i++){ if(i>0 && nums[i]==nums[i-1])continue; int begin=i+1,end=length-1; while(begin<end){ int sum=nums[i]+nums[begin]+nums[end]; if(abs(sum-target)<abs(Min))Min=sum-target; if(sum==target)return target; else if(sum>target)end--; else begin++; } } return Min+target; } private: int abs(int t){ return t>0?t:-t; } void quickSort(vector<int>& nums,int first,int end){ int l=first,r=end,tmp; if(first>=end)return; tmp=nums[l]; while(l<r){ while(l<r && nums[r]>=tmp)r--; if(l<r)nums[l]=nums[r]; while(l<r && nums[l]<=tmp)l++; if(l<r)nums[r]=nums[l]; } nums[l]=tmp; quickSort(nums,first,l-1); quickSort(nums,l+1,end); } };
class Solution: # @param {integer[]} nums # @param {integer} target # @return {integer} def threeSumClosest(self, nums, target): length=len(nums);Min=2147483647 nums.sort() for i in range(length-2): if i>0 and nums[i]==nums[i-1]:continue begin=i+1;end=length-1 while begin<end: sum=nums[i]+nums[begin]+nums[end] if abs(sum-target)<abs(Min):Min=sum-target if sum==target:return target elif sum>target:end-=1 else:begin+=1 return Min+target
LeetCode 16 3Sum Closest(C,C++,Java,Python)
原文地址:http://blog.csdn.net/runningtortoises/article/details/45605493