Question

1822. Sign of the Product of an Array

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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).

 

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

 

Constraints:

• 1 <= nums.length <= 1000
• -100 <= nums[i] <= 100

Solutions

Solution 1: Direct Traversal

The problem requires us to return the sign of the product of the array elements, i.e., return $1$ for positive numbers, $-1$ for negative numbers, and $0$ if it equals $0$.

We can define an answer variable ans, initially set to $1$.

Then we traverse each element $v$ in the array. If $v$ is a negative number, we multiply ans by $-1$. If $v$ is $0$, we return $0$ in advance.

After the traversal is over, we return ans.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

class Solution:
  def arraySign(self, nums: List[int]) -> int:
    ans = 1
    for v in nums:
      if v == 0:
        return 0
      if v < 0:
        ans *= -1
    return ans

class Solution {
  public int arraySign(int[] nums) {
    int ans = 1;
    for (int v : nums) {
      if (v == 0) {
        return 0;
      }
      if (v < 0) {
        ans *= -1;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int arraySign(vector<int>& nums) {
    int ans = 1;
    for (int v : nums) {
      if (!v) return 0;
      if (v < 0) ans *= -1;
    }
    return ans;
  }
};

func arraySign(nums []int) int {
  ans := 1
  for _, v := range nums {
    if v == 0 {
      return 0
    }
    if v < 0 {
      ans *= -1
    }
  }
  return ans
}

impl Solution {
  pub fn array_sign(nums: Vec<i32>) -> i32 {
    let mut ans = 1;
    for &num in nums.iter() {
      if num == 0 {
        return 0;
      }
      if num < 0 {
        ans *= -1;
      }
    }
    ans
  }
}

/**
 * @param {number[]} nums
 * @return {number}
 */
var arraySign = function (nums) {
  let ans = 1;
  for (const v of nums) {
    if (!v) {
      return 0;
    }
    if (v < 0) {
      ans *= -1;
    }
  }
  return ans;
};

int arraySign(int* nums, int numsSize) {
  int ans = 1;
  for (int i = 0; i < numsSize; i++) {
    if (nums[i] == 0) {
      return 0;
    }
    if (nums[i] < 0) {
      ans *= -1;
    }
  }
  return ans;
}