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：跟3sum类似的方法，使用三个指针，让最左边的指针l从[0,size-2]变化，对于每一个固定的最左指针l，中间指针和最右指针在[l+1,size-1]范围内移动，移动方法跟2sum一致。先初始化min_dif=INT_MAX，如果出现更小的dif，则更新min_dif以及closest sum。时间复杂度是O(n^2)，空间复杂度是O(1)。
代码如下：

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        int size = num.size();
        if(size < 3) return 0;
        sort(num.begin(),num.end());
        int min_dif = INT_MAX;
        int closest_sum;
        int l = 0, mid = 1, r = size-1;
        for(;l < size-2;l++){
            mid = l + 1;
            r = size - 1;
            while(mid < r){
            int sum = num[l] + num[mid] + num[r];
            int dif = abs(sum-target);
            if(dif < min_dif){
                closest_sum = sum;
                min_dif = dif;
            }
            if(sum < target) mid++;
            else r--;
            }
        }
        return closest_sum;
    }
};