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)

知道了3 Sum怎么解决，这个就也知道了。只要求出结果就可以了，不需要把结果集再存储到列表中

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
       //如果数组为空或者个数小于三，返回 0
		if (nums == null || nums.length<3) {
			return 0;
		}
		
		//先对数组进行排序
		Arrays.sort(nums);
		//初始化返回的结果
		int res = 0;
		//初始化三个数和与target的最小差值
		int diff = Integer.MAX_VALUE;
		
		
		//与三个数和类似，先定位一个数，再寻找另两个数
		for (int i = 0; i < nums.length-2; i++) {
			if (i > 0 && nums[i] == nums[i-1]) continue;
			int begin = i+1;
			int end = nums.length-1;
			while (begin < end) {
				int threeSum = nums[begin] + nums[end] + nums[i];
				int tempDiff = threeSum - target;//三个数和与target的临时差值
				if (tempDiff == 0) return target;
				if (tempDiff > 0) {
					if (Math.abs(tempDiff) < diff) {//比较临时差值和最小差值
						diff = Math.abs(tempDiff);
						res  = threeSum;
					}
					end--;
				}else if (tempDiff < 0){
					if (Math.abs(tempDiff) < diff) {
						diff = Math.abs(tempDiff);
						res  = threeSum;
					}
					begin++;
				}
				
			}
		}
		return res;
    }
}