Leetcode 1191 K-Concatenation Maximum Sum 动态规划

Leetcode 1191 K-Concatenation Maximum Sum 动态规划

题目描述

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.

例子

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

解题思路

首先，定义新方法maxSum(k)。根据k==1时的解题方法，循环k次即可。但不符合时间复杂度要求。以下是优化方法。
当 len(arr) == 0 时，返回0。
当 k<3 时，可根据k直接求得结果。
当 k<=3 时，记数组arr的总和为sum
当 sum <= 0 时，结果有maxSum(1)和maxSum(2)两种情况;
当 sum > 0 时， 有 maxSum(2) + (k-2)*sum.

Java代码
class Solution {
    public int kConcatenationMaxSum(int[] arr, int k) {
        if ( arr.length == 0 || arr == null ) return 0;
        if (k < 3) return (int) (maxSum(arr, k)%(1e9+7));

        long sum = 0;  for(int num:arr)  sum += num;
        long ans = maxSum(arr, 2);
        return (int) ((ans + (sum>0 ? sum:0) * (k-2))%(1e9+7));
    }
    private long maxSum(int[] arr, int k){
        long sum=0, ans=0;
        for(int i=0; i<k; i++){
            for(int num : arr){
                sum = Math.max(0, sum+=num);
                ans = Math.max(sum, ans);
            }
        }
        return ans;
    }
}

Python代码
class Solution:
    def kConcatenationMaxSum(self, arr: List[int], k: int) -> int:
        def maxSum(arr, res = 0, cur = 0):
            for num in arr:
                res = max(0, res+num)
                cur = max(cur, res)
            return cur
        return ( (k-2)*max(sum(arr), 0) + maxSum(arr*2) )%(10**9+7) if k>1 else maxSum(arr)%(10**9+7)

感谢 花花酱

原文地址：https://www.cnblogs.com/willwuss/p/12241983.html