Maximum Product Subarray

最新推荐文章于 2025-06-06 21:33:25 发布

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最新推荐文章于 2025-06-06 21:33:25 发布

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分类专栏： Dynamic Programming 文章标签：动态规划算法 leetcode

本文链接：https://blog.youkuaiyun.com/u013325815/article/details/106404209

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本文探讨了寻找具有最大乘积的连续子数组的问题，通过动态规划的方法，阐述了解决方案的实现细节。考虑到负数的存在，文章提出了在遇到负数时交换最大值和最小值的策略，以确保正确计算最大乘积。

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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.

思路：这题因为有负数，As we know that on multiplying with negative number max will become min and min will become max, so why not as soon as we encounter negative element, we swap the max and min already. Maximum Non Negative Product in a Matrix 一起看，思路一模一样；

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