152. Maximum Product Subarray - Medium

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Given an integer array nums, find the contiguous subarray within an array (containing at least one number) which has the largest product.

Example 1:

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

Example 2:

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

Overall, we need to loop through the array, at each new element, there‘re 2 options:
u can either multiply the new element to the existing product,
or start a new product from current index (discard previous results)

in this process we need to maintain and update 2 values:
max: the maximum cumulative product UP TO current element starting from SOMEWHERE in the past
min: the minimum cumuliative product UP TO current element
(because when we see a negative number, the "candidate" for max should instead become the previous min product, because a bigger number multiplied by negative becomes smaller)

base case: max = min = res = nums[0], need max and min because max * (negative number) -> min
induction rule:
(1) multiply current element
-> max1 = max * nums[i], max2 = min * nums[i], min1 = min * nums[i], min2 = max * nums[i]
(2) not multiply current element
-> max3 = nums[i], min3 = nums[i]
max = max(max1, max2, max3), min = min(min1, min2, min3), res = max(res, max)

M1: time: O(n), space: O(1)

class Solution {
    public int maxProduct(int[] nums) {
        if(nums == null || nums.length == 0) {
            return 0;
        }
        int max = nums[0];
        int min = nums[0];
        int res = nums[0];
        for(int i = 1; i < nums.length; i++) {
            int tmp = max;
            max = Math.max(Math.max(max * nums[i], min * nums[i]), nums[i]);
            min = Math.min(Math.min(tmp * nums[i], min * nums[i]), nums[i]);
            res = Math.max(res, max);
        }
        return res;
    }
}

152. Maximum Product Subarray - Medium

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原文地址：https://www.cnblogs.com/fatttcat/p/13662431.html