[Leetcode] 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).

跟3SUM一样的想法，排序数组，先选一个数出来，再用target减去这个数，然后用找和为给定值的算法找最小差值，时间复杂度为O(n^2) , 要比枚举所有情况的O(n^3)来的快！

class Solution {
public:
    int getDiff(vector<int> &num, int idx, int target) {
        int low = idx, high = num.size() - 1;
        int res = INT_MAX, tmp;
        while (low < high) {
            tmp = num[low] + num[high] - target;
            res = abs(res) > abs(tmp) ? tmp : res;
            if (tmp > 0) --high;
            else if (tmp < 0) ++low;
            else break;
        }
        return res;
    }
    
    int threeSumClosest(vector<int> &num, int target) {
        int newtarget;
        int diff = INT_MAX, tmp;
        sort(num.begin(), num.end());
        for (int i = 0; i < num.size() - 2; ++i) {
            newtarget = target - num[i];
            tmp = getDiff(num, i + 1, newtarget);
            diff = abs(diff) > abs(tmp) ? tmp : diff;
        }
        return target + diff;
    }
};