154th LeetCode Weekly Contest

A B D均比较简单。先说C题

K-Concatenation Maximum Sum

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

这里有个猜测，

如果K=1，那么我只需要算这个数组的最大子数组和即可。

如果K=2，那么我也只需要算这个数组的最大子数组和，只不过是多加了长度。

如果K>2，那么不需要再讨论什么子数组了，实际上和K<=2类似。所以K>2以后的数组，如果和大于0，则全都要。

class Solution {
public:
    int kConcatenationMaxSum(vector<int>& arr, int k) {
        int sum1 = 0,sum2 = 0;
        int ant = 0;
        int sum3 = 0;
        int arrsize = arr.size();
        for(int i = 0;i<arrsize;i++){
            ant+=arr[i];
            sum3+=arr[i];
            sum1 = max(ant,sum1);
            if(ant<=0){
                ant=0;
            }
        }
        ant = 0;
        for(int i = 0;i<arrsize*2;i++){
            ant+=arr[i%arrsize];
            sum2 = max(ant,sum2);
            if(ant<=0){
                ant=0;
            }
        }
        if(k==1){
            return sum1;
        }
        cout<<sum3<<" "<<sum1<<" "<<sum2<<endl;
        if(sum3>0){
            for(int i=2;i<k;i++){
                sum2 +=sum3;
                sum2%=1000000007;
            }
        }
        return sum2;
    }
};