[LeetCode]22. 3Sum Closest最近三者之和

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).

解法：与3Sum问题一样，可以通过先确定一个数再转化为2Sum问题，只不过此时两个数之和不是等于target，而是三个数之和最接近target，即2Sum问题的两个数之和最接近target-nums[i](i=0,1,...,n-1)。

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int n = nums.size();
        int closest = 0;
        int diff = INT_MAX;
        sort(nums.begin(), nums.end());
        for(int i = 0; i < n - 2; i++)
        {
            int left = i + 1, right = n - 1;
            int tar = target - nums[i];
            while(left < right)
            {
                int tmp = nums[left] + nums[right];
                int dist = abs(tar - tmp);
                if(dist < diff)
                {
                    diff = dist;
                    closest = nums[i] + nums[left] + nums[right];
                }
                if(nums[i] + nums[left] + nums[right] < target)
                    left++;
                else
                    right--;
            }
        }
        return closest;
    }
};

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int n = nums.size();
        int closest = nums[0] + nums[1] + nums[2];
        int diff = abs(closest - target);
        sort(nums.begin(), nums.end());
        for(int i = 0; i < n - 2; i++)
        {
            int left = i + 1, right = n - 1;
            while(left < right)
            {
                int sum = nums[i] + nums[left] + nums[right];
                int dist = abs(sum - target);
                if(dist < diff)
                {
                    diff = dist;
                    closest = sum;
                }
                if(sum < target)
                    left++;
                else
                    right--;
            }
        }
        return closest;
    }
};