# 2 Sum Closest

## Question

> Given an array `nums` of n integers, find two integers in `nums` such that the sum is closest to a given number, `target`.
>
> Return the difference between the sum of the two integers and the target.

Example

Given array `nums = [-1, 2, 1, -4]`, and `target = 4`.

The minimum difference is `1`. (4 - (2 + 1) = 1).

## Analysis

与3 sum closest问题相似，通过先对数组排序，再用两个指针的方法，可以满足 O(nlogn) + O(n) \~ O(nlogn)的时间复杂度的要求

不同的是这里要返回的是diff，所以只用记录minDiif即可。

## Solution

```java
public class Solution {
    /**
     * @param nums an integer array
     * @param target an integer
     * @return the difference between the sum and the target
     */
    public int twoSumCloset(int[] nums, int target) {
        if (nums == null || nums.length < 2) {
            return -1;
        }

        if (nums.length == 2) {
            return target - nums[0] - nums[1];
        }
        Arrays.sort(nums);
        int pl = 0;
        int pr = nums.length - 1;

        int minDiff = Integer.MAX_VALUE;
        while (pl < pr) {
            int sum = nums[pl] + nums[pr];
            int diff = Math.abs(sum - target);
            if (diff == 0) {
                return 0;
            }
            if (diff < minDiff ) {
                minDiff = diff;
            }
            if (sum > target) {
                pr--;
            } else {
                pl++;
            }
        }
        return minDiff;
    }
}
```


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