LeetCode 16. 3Sum Closest

Description:

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.

Example:
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：

首先对数组nums排序，确定第一个数为nums[i]，设定变量begin = i + 1，end = n - 1；
如果当前sum等于target，则begin++，end--；
如果当前sum大于target，则end--；
如果当前sum小于target，则begin++；
然后每次根据sum与target的差值更新result。
注意，result为long类型，避免一开始是INT_MAX最大值，int类型减去一个负数后造成溢出错误。

代码如下：

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