数组求和(两种方法)

#include  <iostream>
using namespace std;

//数组求和

//方法一：时间复杂度为O(n),额外空间为(n)
int ArraySum(int arr[],int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
    {
        sum += arr[i];
    }
    return sum;
}

//方法二：递归的方法，时间复杂度为O(n),额外空间为log(n)
int ArraySum(int arr[], int low, int high)
{
    if (low ==  high)
        return arr[low];
    else if (low<high)
    {
        int mid = (low + high) >> 1;
        return ArraySum(arr, low, mid) + ArraySum(arr, mid+1, high);
    }

}


int main()
{
    int arr[11] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ,11};
    int sum_1, sum_2=0;
    sum_1 = ArraySum(arr, 11);
    sum_2 = ArraySum(arr, 0, 11-1);  //一定要注意，这里是10而不是11，因为数组下标为准，不存在arr[11],已经溢出
    cout << "方法一：" << sum_1 << endl;
    cout << "方法二：" << sum_2 << endl;
    return 0;
}