Maximum Subarray

最新推荐文章于 2025-05-15 16:32:28 发布

转载最新推荐文章于 2025-05-15 16:32:28 发布 · 51 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/kid551/p/4153428.html

本文介绍了一种使用动态规划解决最大子数组和问题的方法。通过定义两个状态变量dp[i]和End[i]，分别表示包含前i个元素的最大子数组和及以第i个元素结尾的最大子数组和。并给出了完整的C++代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

It's obvious an DP problem.

Let's define:

$dp[i] := $ the maximum sum of subarray in the first $i$ elements.

From the base point, the optimal subarray may contain element $A[i]$ or not.

optimal subarray without $A[i]$, so $dp[i] = dp[i-1]$.
optimal subarray ending with $A[i]$.

There're some subtle issues about the second situation. In order to get the second case, we need to know the optimal solution of array ending with $A[i-1]$.

So, we have to define another dp table:

$End[i] := $ the maximum sum of subarray ending with $A[i]$.

Obviously, the $End$ array satisfies:

End[i] = max(End[i-1] + A[i], A[i]);

In fact, we can simplify as

End[i] = (End[i-1] > 0) ? End[i-1] + A[i] : A[i];

As both $dp[i]$ and $End[i]$ only relate the previous step, we can use a variable instead of an array to store the result.

end = (end > 0) ? end + A[i] : A[i];
dp = max(dp, end);

Of course, at the initial stage, we consider only one element $A[0]$, thus, we can initialize $dp, end$ as:

dp = end = A[0];

The complete code is:

class Solution {
public:
    int maxSubArray(int A[], int n) {
        int dp = A[0];
        int end = dp;
        
        for(int i = 1; i < n; ++i){
            end = end > 0 ? end + A[i] : A[i];
            dp = dp > end ? dp : end;
        }
        
        return dp;
    }
};

转载于:https://www.cnblogs.com/kid551/p/4153428.html