LeetCode 1191. K-Concatenation Maximum Sum

最大子数组和在重复数组中的求解

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本文探讨了在一个整数数组被重复k次后的最大子数组和的求解方法，考虑到子数组长度可以为0的情况，返回结果需进行模运算。通过三个实例演示了算法的具体应用，并在约束条件下提供了Python实现代码。

Given an integer array arr and an integer k, modify the array by repeating it k times.

For example, if arr = [1, 2] and k = 3 then the modified array will be [1, 2, 1, 2, 1, 2].

Return the maximum sub-array sum in the modified array. Note that the length of the sub-array can be 0 and its sum in that case is 0.

As the answer can be very large, return the answer modulo 10^9 + 7.

Example 1:

Input: arr = [1,2], k = 3
Output: 9

Example 2:

Input: arr = [1,-2,1], k = 5
Output: 2

Example 3:

Input: arr = [-1,-2], k = 7
Output: 0

Constraints:

1 <= arr.length <= 10^5
1 <= k <= 10^5
-10^4 <= arr[i] <= 10^4

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反应有些慢，如果3个拼接比2个拼接的结果大，说明中间一定用满了。。。

class Solution:
    def max_subarry(self,arr):
        res,pre = 0,0
        for num in arr:
            pre = pre+num if pre>0 else num
            res = max(pre,res)
        return res
        
    def kConcatenationMaxSum(self, arr: List[int], k: int) -> int:
        m1 = self.max_subarry(arr)
        mod = 10**9+7
        if (k == 1):
            return m1%mod
        m2 = self.max_subarry(arr*2)
        if (m1 == m2):
            return m1%mod
        total = sum(arr)
        if (total > 0):
            return (m2+total*(k-2))%mod
        else:
            return m2%mod