Maximum Product Subarray

描述

Find the contiguous subarray within an array (containing at least one number) which has the largest product.

For example, given the array [2,3,-2,4], the contiguous subarray [2,3] has the largest product = 6.

分析

这题跟“最大连续子序列和”非常类似，只不过变成了“最大连续子序列积”，所以解决思路也很类似。

仅仅有一个小细节需要注意，就是负负得正，两个负数的乘积是正数，因此我们不仅要跟踪最大值，还要跟踪最小值。

动规

Python

// maximum-product-subarray
// 时间复杂度O(n)，空间复杂度O(1)
public class Solution {
    public int maxProduct(int[] nums) {
        int maxLocal = nums[0];
        int minLocal = nums[0];
        int global = nums[0];

        for(int i = 1; i < nums.length; i++){
            int temp = maxLocal;
            maxLocal = Math.max(Math.max(nums[i] * maxLocal, nums[i]), nums[i] * minLocal);
            minLocal = Math.min(Math.min(nums[i] * temp, nums[i]), nums[i] * minLocal);
            global = Math.max(global, maxLocal);
        }
        return global;
    }
}

Java

// maximum-product-subarray
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int maxLocal = nums[0];
        int minLocal = nums[0];
        int global = nums[0];

        for(int i = 1; i < nums.size(); i++){
            int temp = maxLocal;
            maxLocal = max(max(nums[i] * maxLocal, nums[i]), nums[i] * minLocal);
            minLocal = min(min(nums[i] * temp, nums[i]), nums[i] * minLocal);
            global = max(global, maxLocal);
        }
        return global;
    }
};

C++

# maximum-product-subarray
# 时间复杂度O(n)，空间复杂度O(1)
def maxProduct(nums):
    maxLocal = nums[0]
    minLocal = nums[0]
    global_max = nums[0]

    for i in range(1, len(nums)):
        temp = maxLocal
        maxLocal = max(max(nums[i] * maxLocal, nums[i]), nums[i] * minLocal)
        minLocal = min(min(nums[i] * temp, nums[i]), nums[i] * minLocal)
        global_max = max(global_max, maxLocal)

    return global_max

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