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).

tag: array, two pointers

有关“两个指针”的问题，基本思想是将一个指针置于数组头，一个指针置于数组尾，然后两个指针向中间靠拢。

本题也是这样。

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