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).

算法：固定第一个，另外两个数适用双指针。O(N^2)

java：

public class Solution {
    public int threeSumClosest(int[] num, int target) {
        int len=num.length;
        
        int distance=Integer.MAX_VALUE;
        int sum=0;
        Arrays.sort(num);
        
        for(int i=0;i<len-2;i++){
            int s=i+1;
            int e=len-1;
            while(s<e){
                int cnt= num[i]+num[s]+num[e];
                int dis = Math.abs(target-cnt);
               
                if(dis<=distance){
                    sum=cnt;
                    distance=dis;
                    if(sum==target){
                        return sum;
                    }else if(sum<target){
                        s++;
                    }else {
                        e--;
                    }
                }else if(cnt<target){
                    s++;
                }else{
                    e--;
                }
            }
        }
        return sum;
    }
}

c++:

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

3Sum Closest