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).

/*
The naive way is to search every three-element pair and define a variable to store the sum of the pair which is closest to the target.
time:O(n^3), space:O(1)

The better way is to use two pointers. Like 3Sum.
The variable 'closestSum' can not be initialized to Integer.MAX_VALUE, for 'Math.abs(closestSum - target)' will out of range when the target is negative. 
time:O(n^2), space:O(1)
*/

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        if (nums == null || nums.length < 3) {
            return Integer.MAX_VALUE;
        }
        Arrays.sort(nums);
        int closestSum = nums[0] + nums[1] + nums[2];
        for(int i = 0; i < nums.length - 2; i++) {
            int left = i + 1;
            int right = nums.length - 1;
            while(left < right){
                int sum = nums[i] + nums[left] + nums[right];
                if (sum == target) {
                    return target;
                } else if (sum < target) {
                    left++;
                } else {
                    right--;
                }
                if (Math.abs(sum - target) < 
                    Math.abs(closestSum - target)) {
                    closestSum = sum;
                }
            }
        } 
        return closestSum;
    }
}