Maximum Product Subarray 求最大子数组乘积

最新推荐文章于 2025-03-05 00:18:42 发布

原创最新推荐文章于 2025-03-05 00:18:42 发布 · 296 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #最大子数组乘积

lintcode 专栏收录该内容

4 篇文章

订阅专栏

本文探讨了如何在整数数组中找到具有最大乘积的连续子数组，提供了一个动态规划解决方案，通过维护最大和最小产品子数组来解决负数带来的挑战。

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.

以下摘自OJ官方解答，大体思路相同，写法简洁：

Besides keeping track of the largest product, we also need to keep track of the smallest product. Why? The smallest product, which is the largest in the negative sense could become the maximum when being multiplied by a negative number.

Let us denote that:

f(k) = Largest product subarray, from index 0 up to k.

Similarly,

g(k) = Smallest product subarray, from index 0 up to k.

Then,

f(k) = max( f(k-1) * A[k], A[k], g(k-1) * A[k] )
g(k) = min( g(k-1) * A[k], A[k], f(k-1) * A[k] )

There we have a dynamic programming formula. Using two arrays of size n, we could deduce the final answer as f(n-1). Since we only need to access its previous elements at each step, two variables are sufficient.

public int maxProduct(int[] A) {
   assert A.length > 0;
   int max = A[0], min = A[0], maxAns = A[0];
   for (int i = 1; i < A.length; i++) {
      int mx = max, mn = min;
      max = Math.max(Math.max(A[i], mx * A[i]), mn * A[i]);
      min = Math.min(Math.min(A[i], mx * A[i]), mn * A[i]);
      maxAns = Math.max(max, maxAns);
   }
   return maxAns;
}