# 16. 3Sum Closest

## 1.問題

![](https://901207480-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LGKoChvN9am4__HCIRK%2F-LLxK83apAqmrnEf-HeA%2F-LLxM27wV2A_o36i1Q85%2F%E8%9E%A2%E5%B9%95%E5%BF%AB%E7%85%A7%202018-09-09%20%E4%B8%8B%E5%8D%884.03.51.png?alt=media\&token=9d1fcc59-a408-418e-a434-d02ea117fa2c)

## 2.想法

* 提問
* function header, parameter
* test input
* 說明想法
  * 對序列後排序
  * 使用兩個迴圈
  * 分別使用3個指標,  依題意的目的是要計算總和與目標值的差值
    * 若差值 == 0, 則記錄i, j , k的數值
    * 若差值 < 0, 則表示總和太小, 讓j++
    * 若差值 > 0, 則表示總和太大, 讓k--
* 測試計算複雜度: O(n^2) ?

## **3.程式碼**

```
class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        vector<vector<int>> res;
        int size = nums.size();
        if (size < 3) {
            return 0;
        }
        int value = INT_MAX, minSum = -1;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < size; i++) {
            int j = i + 1;
            int k = size - 1;
            while (j < k) {
                int sum = nums[i] + nums[j] + nums[k];
                int diff = sum - target;
                if (diff == 0){
                    return sum;
                } else if (diff > 0) {
                    k--;
                } else {
                    j++;
                }
                
                if (abs(diff) < value) {
                    value = abs(diff);
                    minSum = sum;
                }
            }
        }
        
        return minSum;
    }
};
```