40-3Sum Closest

1. 3Sum Closest My Submissions QuestionEditorial Solution
   
   Total Accepted: 76185 Total Submissions: 262100 Difficulty: Medium
   
   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).

Submission Details

120 / 120 test cases passed.

Status: Accepted

Runtime: 24 ms

beats:25.39%

思路：参看我写的3sum，4sum，思路差不多，

先固定一个数，剩下的两数进行左右夹逼，跳过不必要的判断，

用best_sum来记录距离最近的和，用error记录距离target的距离

#define IMAX numeric_limits<int>::max()
class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        size_t n = nums.size();
        sort(nums.begin(),nums.end());
        int best_sum=0,error=IMAX;
        for(int i=0;i<n;++i){
            int beg = i+1,end = n-1;
            while(beg<end){
                int tsum =nums[i]+nums[beg]+nums[end];
                if(tsum==target) {
                        best_sum = tsum;
                        while(++beg<end&&nums[beg]==nums[beg-1]);//如果和前一个数一样跳过
                        while(--end>beg&&nums[end]==nums[end]);//如果和后一个数一样，跳过
                }
                else{
                        if(tsum<target)beg++;
                        else end--;
                }
                if(abs(target-tsum)<error){
                        error = abs(target-tsum);
                        best_sum = tsum;
                }
            }
        }
        return best_sum;
    }
};