There is a function signFunc(x)
that returns:
1
ifx
is positive.-1
ifx
is negative.0
ifx
is equal to0
.
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
Solution: Sign Only
No need to compute the product, only track the sign changes. Flip the sign when encounter a negative number, return 0 if there is any 0 in the array.
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int arraySign(vector<int>& nums) { int sign = 1; for (int x : nums) { if (x == 0) return 0; else if (x < 0) sign = -sign; } return sign; } }; |