---
# System prepended metadata

title: ' Leetcode [No. 2485] Find the Pivot Integer (EASY) 解題心得'
tags: [leetcode]

---

# Leetcode [No. 2485] Find the Pivot Integer (EASY) 解題心得

## 題目

https://leetcode.com/problems/find-the-pivot-integer/description/

## 思路
+ 這個解法就是用兩個陣列去紀錄每一個integer前後之和，如果preSum==postSum，就代表這個i是pivot。
```c++=
class Solution {
public:
    int pivotInteger(int n) {
        vector<int> preSum(n+1,0);
        vector<int> postSum(n+1,0);
        // calculate preSum
        int pre = 0, post = 0;
        for (int i = 0 ; i <= n ; i++)
        {
            preSum[i] = pre;
            pre += i;
        }
        for (int i = n ; i >= 0; i--)
        {
            postSum[i] = post;
            post+= i;
        }

        for (int i = 0 ; i <= n; i++)
        {
            // cout << preSum[i] << "\t" << postSum[i] << endl;
            if(preSum[i] == postSum[i]) return i;
        }
        return -1;
    }
};
```

### 解法分析
+ time complexity: O(n)


### 執行結果

![image](https://hackmd.io/_uploads/BJZDAi0pp.png)
