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# 15. 3Sum

## 1.問題

![](/files/-LLxJuq-OI48P0O48IFh)

## 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,  依題意的目的是要總和為0,  因此必須要有正負數
    * i 所指向的必須是負數, 因此若nums\[i] > 0則break
    * 若總和 == 0, 則記錄i, j , k的數值
    * 若總和 < 0, 則表示總和太小, 讓j++
    * 若總和 > 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:
    vector<vector<int>> threeSum(vector<int>& num) {
        vector<vector<int>> res;
        int size = num.size();
        if (size < 3) {
            return res;
        }
        sort(num.begin(), num.end());
        
        for (int i = 0; i < size - 2; i++) {
            int a = num[i];
            if (a > 0) {
                break;
            }
            if (i > 0 && a == num[i -1]){
                continue;
            }
            int j = i + 1;
            int k = size - 1;
            while (j < k) {
                int b = num[j];
                int c = num[k];
                int sum = a + b + c;
                if (sum == 0){
                    res.push_back({a, b, c});
                    while(b == num[++j]);
                    while(c == num[--k]);
                } else if (sum > 0){
                    k--;
                } else {
                    j++;
                }
            }
        }
        return res;
    }
};
```


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