Question

这是我的代码：

long sum(int[] arr) {
    long sum=0;
    int j,k;
    long sumtoadd;
    for (int i = 0; i < arr.length; i++)
    {
        for (j = i; j < arr.length; j++)
        {
            sumtoadd = 0;
            for (k = i; k <= j; k++)
            {
                sumtoadd = sumtoadd + arr[k];
            }
            sum = sumtoadd + sum;
        }
    }
    return sum;
}

示例：

数组：{1,2,3}输出：20

数组：{1,1,1}输出：10

我试图找到数组的所有连续子数组的总和，但是对于某些情况，时间已经超过了。 此解决方案适用于除大型机箱外的所有情况。有更好的解决方案吗？

Answer 1

public class Test1 {

static long sum2(int[] arr) {
    long n = arr.length;
    long sum = 0;
    for (int i = 0; i < n; i++) {
        sum += (n-i)*(i+1)*arr[i];
    }
    return sum;
}

static int[] arr1 = new int[]{1,2,3,4,5,6,7,8,9};
static int[] arr2 = new int[]{1,1,1,1};

public static void main(String[] args) {
    System.out.println("sum(arr1) = " + sum(arr1));
    System.out.println("sum2(arr1) = " + sum2(arr1));
    System.out.println("sum(arr2) = " + sum(arr2));
    System.out.println("sum2(arr2) = " + sum2(arr2));
}

//your code to check
static long sum(int[] arr) {
    long sum=0;
    int j,k;
    long sumtoadd;
    for (int i = 0; i < arr.length; i++)
    {
        for (j = i; j < arr.length; j++)
        {
            sumtoadd = 0;
            for (k = i; k <= j; k++)
            {
                sumtoadd = sumtoadd + arr[k];
            }
            sum = sumtoadd + sum;
        }
    }
    return sum;
}

}

Answer 2

而不是三个嵌套循环，可以在N数组元素的一个循环中找到答案。

推理：

如果给定的数组元素A[i]出现在K个子数组中，那么它会将其值贡献给K个子数组，这意味着它贡献K - 它的总值乘以A[i]和。现在元素0出现在所有子数组中，从i到i+1（包括）开始，从N到K = (i+1)*(N-i)（包括）结束，所以A[i]。

摘要：每个(i+1)*(N-i)*A[i]贡献class Room(object): """Create general features of a general room Each room has a certain capacity depending on its type(an office or a living room) """ def __init__(self): super(Room, self).__init__() self.capacity = ''; class Office(Room): """ Add specific features to make a room an office. An office has a name and a space for maximum of 4 people. """ def __init__(self, rname): #create object if name does not start with a digit if not (rname[0].isdigit()): super(Office,self).__init__() self.name = rname #office name self.capacity = 4 #number of spaces per office。在一个循环中完成，然后你就完成了。

最少时间内数组的连续子数组的总和

2 个答案: