# 【LeetCode】 1822. Sign of the Product of an Array

## Description

> There is a function `signFunc(x)` that returns:

> * `1` if `x` is positive.
> * `-1` if `x` is negative.
> * `0` if `x` is equal to `0`.

> You are given an integer array `nums`. Let `product` be the product of all values in the array `nums`.
> Return `signFunc(product)`.

> Constraints:
> * `1 <= nums.length <= 1000`
> * `-100 <= nums[i] <= 100`

> 這裡有一個函式 `signFunc(x)` 回傳：

> * 如果 `x` 是正數回傳 `1`。
> * 如果 `x` 是負數回傳 `-1`。
> * 如果 `x` 等於 `0` 回傳 `0`。

> 你會收到一個整數陣列 `nums`。讓 `product` 成為所有陣列 `nums` 值的乘積。
> 回傳 `signFunc(product)`。

> 限制：
> * `1 <= nums.length <= 1000`
> * `-100 <= nums[i] <= 100`

## Example:

Example 1:

Input: nums = [-1,-2,-3,-4,3,2,1]
Output: 1
Explanation: The product of all values in the array is 144, and signFunc(144) = 1

Example 2:

Input: nums = [1,5,0,2,-3]
Output: 0
Explanation: The product of all values in the array is 0, and signFunc(0) = 0

Example 3:

Input: nums = [-1,1,-1,1,-1]
Output: -1
Explanation: The product of all values in the array is -1, and signFunc(-1) = -1

## Solution

* 乘積只會受到正數負數影響，因此不用考慮數值大小
* 用一個變數紀錄目前乘積的正負，遍歷一次陣列
* 只要遇到 `0` 乘積必定為 `0`
* 負號的話則是將正負相反

### Code

class Solution {
public:
    int arraySign(vector<int>& nums) {
        int sign = 1;
        for(int i = 0; i < nums.size(); i++)
        {
            if(nums[i] == 0)
                return 0;
            else if(nums[i] < 0)
                sign *= -1;
        }
        return sign;
    }
};

###### tags: `LeetCode` `C++`