Python
class Solution:
def maxProduct(self, nums: List[int]) -> int:
n = len(nums)
max_res = nums[0]
min_res = nums[0]
res = nums[0]
for i in range(1, n):
tmp_max, tmp_min = max_res, min_res
max_res = max(nums[i], max(nums[i] * tmp_max, nums[i] * tmp_min))
min_res = min(nums[i], min(nums[i] * tmp_max, nums[i] * tmp_min))
res = max(max_res, res)
return res
Java
法1:DP
参考:https://blog.csdn.net/Innocence02/article/details/128326633
f[i]表示以i结尾的连续子数组的最大乘积,d[i]表示以i结尾的连续子数组的最小乘积 。
如果只有正数,我们只需要考虑最大乘积f[i];有负数,需要考虑与负数相乘的数,越小越好。所以d[i] 维护最小乘积。
class Solution {
public int maxProduct(int[] nums) {
int max = nums[0], min = nums[0], res = nums[0];
for (int i = 1; i < nums.length; ++i) {
int tmpMax = max, tmpMin = min;
max = Math.max(nums[i], Math.max(tmpMax * nums[i], tmpMin * nums[i]));
min = Math.min(nums[i], Math.min(tmpMax * nums[i], tmpMin * nums[i]));
res = Math.max(max, res);
}
return res;
}
}