[LeetCode]—3Sum Closest 求数组中三个数之和最接近给定target的组合

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).

思路1：最容易想到，采用和3Sum相同的方法，先排序，再查找，记录下每个组合的和，找出最接近target的组合。但是显然时间复杂度较高，超时。

class Solution{
public:
    int threeSumClosest(vector<int> &num,int target){
    int result=INT_MAX;
    sort(num.begin(),num.end());
    
    for(vector<int>::iterator a=num.begin();a<num.end()-2;a=upper_bound(a,num.end()-2,*a)){
            for(vector<int>::iterator b=a+1;b<num.end()-1;b=upper_bound(b,num.end()-1,*b)){
                for(vector<int>::iterator c=b+1;c<num.end();c=upper_bound(c,num.end(),*c)){
              if(abs(*a+*b+*c-target)<result)result=*a+*b+*c;
                    }
                }
            }

        return result;
    }
    

};

思路二：先排序是必须的。然后由于要求找最接近的，那么从两边向其逼近是最快的，可以省略单方向很多无谓的比较。采用第2，3 个数分别从两端开始向中间夹逼

 在b<c的情况下：        

                如果a+b+c的sum>target 那么 c--

                如果a+b+c的sum<target 那么 b++

                如果a+b+c的sum=target  那么 return      

                                      

class Solution{
public:
  int threeSumClosest(vector<int> &num,int target){
    int result=0;
    int min_gap=INT_MAX;

    sort(num.begin(),num.end());
   
    if(num.size()==3){
     return num[0]+num[1]+num[2];
    }

    for(vector<int>::iterator a=num.begin();a<num.end()-2;a=upper_bound(a,num.end()-2,*a)){
         vector<int>::iterator c=num.end()-1;
         vector<int>::iterator b=a+1;
         
         while(b<c){      // b!=c
            int sum=*a+*b+*c;
            int gap=abs(sum-target);
            if(sum<target)b=upper_bound(b,num.end()-1,*b);
            if(sum>target)c=lower_bound(b,c,*c)-1;
            if(sum==target) return sum;    
           if(gap<min_gap){
              result=sum;
              min_gap=gap;
           }
        }        
    }

           return result;
    }
    

};