16. 3Sum Closest(中等)

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).

如上例， -1 + 1 = 0 不也接近 1 吗？不行，因为人家要求３数之和．
思路同15题 3Sum. 既固定i, 让 lo = i + 1; hi = len - 1;
关键是对sum == ta, sum > ta, sum < ta的处理，以及退出while(lo < hi)之后的处理．

人家想法，咱代码：
(O(n^2)) time, (O(1)) extra space.

int threeSumClosest(vector<int>& A, int ta) {
	const int n = A.size();
	int min = INT_MAX; // 存放最接近ta的三个数和
	sort(A.begin(), A.end());

	for (int i = 0; i < n - 2; i++) {
		// 剔除连续值
		if (i > 0 && A[i] == A[i - 1]) continue;
		int lo = i + 1, hi = n - 1;
		
		// lo==hi 不可以，sum = A[i] + A[lo] + A[hi]会出问题
		while (lo < hi) {
			int sum = A[i] + A[lo] + A[hi];

			// 保存最接近ta的３数之和
			min = abs(sum - ta) < abs(min) ? (sum - ta) : min;

			if (sum == ta) return ta;
			else if (sum > ta) hi--;
			else lo++;
		}
	}
	return min + ta; // min = sum - ta, we ask sum now!
}