算法: 最大乘积子数组152. Maximum Product Subarray

Given an integer array nums, find a contiguous non-empty subarray within the array that has the largest product, and return the product.

The test cases are generated so that the answer will fit in a 32-bit integer.

A subarray is a contiguous subsequence of the array.

Example 1:

Input: nums = [2,3,-2,4]
Output: 6
Explanation: [2,3] has the largest product 6.

Example 2:

Input: nums = [-2,0,-1]
Output: 0
Explanation: The result cannot be 2, because [-2,-1] is not a subarray.

Constraints:

• 1 <= nums.length <= 2 * 104
• -10 <= nums[i] <= 10
• The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

1. 左右两个方向，计算乘积，如果遇到0，则乘以1。java解法

 public int maxProduct(int[] A) {
        int n = A.length, res = A[0], l = 0, r = 0;
        for (int i = 0; i < n; i++) {
            l =  (l == 0 ? 1 : l) * A[i];
            r =  (r == 0 ? 1 : r) * A[n - 1 - i];
            res = Math.max(res, Math.max(l, r));
        }
        return res;
    }

2. python解法，直接翻转数组

class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        B = nums[::-1]
        for i in range(1, len(nums)):
            if nums[i-1] != 0: nums[i] *= nums[i-1]
            if B[i-1] != 0: B[i] *= B[i-1]
        return max(nums + B)

更加简洁的实现

class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        B = nums[::-1]
        for i in range(1, len(nums)):
            nums[i] *= nums[i-1] or 1
            B[i] *= B[i-1] or 1
        return max(nums + B)

解释 A[i] *= A[i - 1] or 1
它就像这样(5 or 1)= 5 逻辑加法，如果是(x or 1)= x 一直如此，除非 x 为零，那么它是 1

参考

https://leetcode.com/problems/maximum-product-subarray/discuss/183483/JavaC%2B%2BPython-it-can-be-more-simple