3Sum Closest

Given an arraySofnintegers, find three integers inSsuch 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).

這道題和3Sum差不多，不過也有不一樣的，主要是：

1 這裡不用判斷處理重複問題

2 要比較其中的三個數的和與目標數的差的大小。

class Solution {
public:
	int threeSumClosest(vector<int> &num, int target)
	{
		switch (num.size())
		{
		case 0: 
			return 0;
		case 1:
			return num[0];
		case 2:
			return num[0] + num[1];
		default:
			break;
		}
		sort(num.begin(), num.end());
		int closet = 0;
		int sum = 0;
		int i = 0, j = 0, k = num.size()-1;
		int diff = INT_MAX;

		for (i = 0; i < k-1; i++)
		{
			for (j = i+1; j < k;)
			{
				sum = num[i] + num[j] + num[k];
				if (sum == target)
				{
					return sum;
				}
				if (abs(sum-target) < diff)
				{
					closet = sum;
					diff = abs(sum - target);
				}
				if (sum < target)
				{
					j++;
				}
				else
				{
					k--;
				}
			}
		}
		return closet;
	}
};
 

//2014-1-25 update8.24
class Solution125 {
public:
	int threeSumClosest(vector<int> &num, int target)
	{
		int closest = INT_MAX;
		int res = 0;
		sort(num.begin(), num.end());

		for (int i = 0; i < num.size(); i++)
		{
			for (int j = i+1, k = num.size()-1; j < k; )
			{
				int sum = num[i]+num[j]+num[k];
				if (sum == target) return sum;
				int t = abs(sum-target);
				if (t<closest)
				{
					res = sum;
					closest = t;
				}
				if (sum < target) j++;
				else if (sum > target) k--;
			}
			while (i < num.size() && num[i] == num[i+1]) i++;
		}
		return res;
	}
};