Link: https://leetcode.cn/problems/maximum-length-of-subarray-with-positive-product/
Question
difficulty: mid
adj diff:
Given an array of integers nums, find the maximum length of a subarray where the product of all its elements is positive.
A subarray of an array is a consecutive sequence of zero or more values taken out of that array.
Return the maximum length of a subarray with positive product.
Example 1:
Input: nums = [1,-2,-3,4]
Output: 4
Explanation: The array nums already has a positive product of 24.
Example 2:
Input: nums = [0,1,-2,-3,-4]
Output: 3
Explanation: The longest subarray with positive product is [1,-2,-3] which has a product of 6.
Notice that we cannot include 0 in the subarray since that'll make the product 0 which is not positive.
Example 3:
Input: nums = [-1,-2,-3,0,1]
Output: 2
Explanation: The longest subarray with positive product is [-1,-2] or [-2,-3].
Constraints:
1 <= nums.length <= 105
-109 <= nums[i] <= 109
重点:正数不用管,负数 count 一下。
Code
class Solution {
public int getMaxLen(int[] nums) {
int len = nums.length;
int firstPos = -1; // careful here!!!
int firstNeg = -1;
int countNeg = 0;
int maxLen = 0;
for (int i = 0; i < len; i++) {
if (nums[i] == 0) {
firstPos = i; // careful here!!!
firstNeg = -1;
countNeg = 0;
} else if (nums[i] < 0) {
countNeg++;
if (firstNeg == -1) {
firstNeg = i;
}
}
// check length (product is positive)
if (countNeg % 2 == 1) {
// odd number of '-', thus need to remove a negative
maxLen = Math.max(maxLen, i - firstNeg);
} else {
maxLen = Math.max(maxLen, i - firstPos);
}
}
return maxLen;
}
}