Maximum Product Subarray

最新推荐文章于 2019-09-26 09:33:48 发布

原创最新推荐文章于 2019-09-26 09:33:48 发布 · 204 阅读

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本文介绍了一种寻找数组中连续子数组的最大乘积的方法。通过维护局部最大值和局部最小值来解决负数可能反转最大最小值的问题。该算法能够高效地找到包含至少一个元素的连续子数组的最大乘积。

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Maximum Product Subarray

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.

We need to keep a local max and a local min, because the local min may become local max if it is multiplied by a negative number. We also need to keep a global max variable to represent the current global max.

public class Solution {
    public int maxProduct(int[] nums) {
        if (nums == null || nums.length == 0)
            return Integer.MIN_VALUE;
        
        int max_local = nums[0];
        int min_local = nums[0];
        int max_global = nums[0];
        
        for (int i = 1; i < nums.length; i++) {
            int max_copy = max_local;
            max_local = Math.max(Math.max(nums[i], nums[i] * max_local), min_local * nums[i]);
            min_local = Math.min(Math.min(nums[i], nums[i] * max_copy), min_local * nums[i]);
            max_global = Math.max(max_local, max_global);
        }
        
        return max_global;
    }
}