[LeetCode] 16. 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).

解法： 

　　这道题让我们求最接近给定值的三数之和，是在之前那道3Sum的基础上又增加了些许难度，那么这道题让我们返回这个最接近于给定值的值，即我们要保证当前三数和跟给定值之间的差的绝对值最小，所以我们需要定义一个变量diff用来记录差的绝对值，然后我们还是要先将数组排个序，然后开始遍历数组，思路跟那道三数之和很相似，都是先确定一个数，然后用两个指针left和right来滑动寻找另外两个数，每确定两个数，我们求出此三数之和，然后算和target的差的绝对值，如果比diff小则更新diff的值。注意，如果三数之和正好等于target，直接返回target即可，因为不会有其他数比本身更接近自己了。代码如下：

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        int diff = nums[0] + nums[1] + nums[2] - target;
        
        Arrays.sort(nums);
        
        for (int i = 0; i < nums.length - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            int left = i + 1, right = nums.length - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                diff = Math.abs(sum - target) < Math.abs(diff) ? sum - target : diff;
                
                if (sum == target) {
                    return sum;
                } else if (sum < target) {
                    while (left < right && nums[left++] == nums[left]) {}
                } else {
                    while (left < right && nums[right--] == nums[right]) {}
                }
            }
        }
        return target + diff;
    }
}