[LeetCode] 16. 3Sum Closest

[Grand Yang]

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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:

**Input:** nums = [-1,2,1,-4], target = 1
**Output:** 2
**Explanation:** The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

Example 2:

**Input:** nums = [0,0,0], target = 1
**Output:** 0
**Explanation:** The sum that is closest to the target is 0. (0 + 0 + 0 = 0). 

Constraints:

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

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，那么这道题让返回这个最接近于给定值的值，即要保证当前三数和跟给定值之间的差的绝对值最小，所以需要定义一个变量 diff 用来记录差的绝对值，然后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，求出此三数之和，然后算和给定值的差的绝对值存在 newDiff 中，然后和 diff 比较并更新 diff 和结果 closest 即可，代码如下：

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int closest = nums[0] + nums[1] + nums[2], n = nums.size();
        int diff = abs(closest - target);
        sort(nums.begin(), nums.end());
        for (int i = 0; i < n - 2; ++i) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            int left = i + 1, right = n - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                int newDiff = abs(sum - target);
                if (diff > newDiff) {
                    diff = newDiff;
                    closest = sum;
                }
                if (sum < target) ++left;
                else --right;
            }
        }
        return closest;
    }
};

Github 同步地址：

#16

类似题目：

3Sum Smaller

3Sum

参考资料：

https://leetcode.com/problems/3sum-closest/

https://leetcode.com/problems/3sum-closest/discuss/7883/C%2B%2B-solution-O(n2)-using-sort

https://leetcode.com/problems/3sum-closest/discuss/7872/Java-solution-with-O(n2)-for-reference

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[lld2006]

可以优化一下， 当nums[i] *3 > target 的时候直接return nums[i]+nums[i+1]+nums[i+2] 和当前closest中最接近的那个

[Grand Yang]

已添加，多谢指出～

[zhaodonglin]

if (nums[i] * 3 > target) return min(closest, nums[i] + nums[i + 1] + nums[i + 2]);
should be like this:
if (nums[i] * 3 > target) {
total = nums[i] + nums[i + 1] + nums[i + 2];
if (abs(total-target) < diff) { return total;}
else{return closest}
}