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.

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).

分析：同样采取i,j,k三个对数组扫描，记录target closet to target;

code:

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        int subnum=num[0]+num[1]+num[2];
        
        int sum;
        sort(num.begin(),num.end());
        for(int i=0;i<num.size();++i)
        {
            int j=i+1;
            int k=num.size()-1;
            while(j<k)
            {
                sum=num[i]+num[j]+num[k];
                if(sum==target)
                    return target;
                else if(sum>target)
                {
                    if(abs(sum-target)<abs(subnum-target))
                        subnum=sum;
                    --k;
                }
                else
                {
                    if(abs(sum-target)<abs(subnum-target))
                        subnum=sum;
                    ++j;
                }
            }
            
            int tmp=abs(sum-target);
            if(tmp<abs(subnum-target))
                subnum=sum;
        }
        return subnum;
    }
};

上面的程序比较繁琐，可以简化

code：

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        int subnum=num[0]+num[1]+num[2];
        
        int sum;
        sort(num.begin(),num.end());
        for(int i=0;i<num.size();++i)
        {
            int j=i+1;
            int k=num.size()-1;
            while(j<k)
            {
                sum=num[i]+num[j]+num[k];
                if(abs(sum-target)<abs(subnum-target))
                        subnum=sum;
                
                if(sum>target)
                    --k;
                else
                    ++j;
            }
        }
        return subnum;
    }
};