# 16. 3Sum Closest

**16。3Sum Closest**

Given an integer array `nums` of length `n` and an integer `target`, find three integers in `nums` such that the sum is closest to `target`.

Return *the sum of the three integers*.

You may assume that each input would have exactly one solution.

**Example 1:**

<pre><code><strong>Input: nums = [-1,2,1,-4], target = 1
</strong><strong>Output: 2
</strong><strong>Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: nums = [0,0,0], target = 1
</strong><strong>Output: 0
</strong><strong>Explanation: The sum that is closest to the target is 0. (0 + 0 + 0 = 0).
</strong></code></pre>

**Constraints:**

* `3 <= nums.length <= 500`
* `-1000 <= nums[i] <= 1000`
* `-104 <= target <= 104`

***My Solutions:***

和#15类似，先sort数组，然后用two pointers的思想，对数组里的每个数字除了ta本身（index=i），用i+1当left，nums.length-1当right，计算加起来的结果。

Time: O(N^2)

```
class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int res = Integer.MAX_VALUE;
        for (int i = 0; i < nums.length - 2; i++) {
            if (i == 0 || (i > 0 && nums[i] != nums[i - 1])) {
                int l = i + 1, r = nums.length - 1;
                while (l < r) {
                    int sum = nums[i] + nums[l] + nums[r];
                    if (sum > target) r--;
                    else l++;
                    if (Math.abs(sum - target) < Math.abs(res - target)) res = sum;
                }
            }
        }
        return res;
    }
}
```