标签:乘积 nums int neg pos else maxl 正数 LeetCode
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.
状态转移方程
1.nums[i] > 0
pos[i] = pos[i-1] + 1;
neg[i] = neg[i-1] == 0 ? 0 : neg[i-1] + 1
2.nums[i] < 0
neg[i] = pos[i-1] + 1;
pos[i] = neg[i-1] == 0 ? 0 : neg[i-1] + 1;
3.nums[i] == 0
neg[i] = 0;
pos[i] = 0;
边界条件
if(nums[0] > 0) pos[0] = 1
else neg[0] = 1
代码
class Solution {
public:
int getMaxLen(vector<int>& nums) {
int len = nums.size();
int pos[100005] = {0}, neg[100005] = {0};
if(nums[0] > 0)
pos[0] = 1;
else if(nums[0] < 0)
neg[0] = 1;
int maxl = pos[0];
for(int i = 1; i < len; i++)
{
if(nums[i] < 0)
{
neg[i] = pos[i-1] + 1;
pos[i] = neg[i-1] == 0 ? 0 : neg[i-1] + 1;
}
else if(nums[i] == 0)
{
neg[i] = 0;
pos[i] = 0;
}
else if(nums[i] > 0)
{
pos[i] = pos[i-1] + 1;
neg[i] = neg[i-1] == 0 ? 0 : neg[i-1] + 1;
}
if(pos[i] > maxl)
maxl = pos[i];
}
return maxl;
}
};
标签:乘积,nums,int,neg,pos,else,maxl,正数,LeetCode 来源: https://blog.csdn.net/ucler/article/details/119210891
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