📜  数组平衡索引的C程序

📅  最后修改于: 2022-05-13 01:55:28.733000             🧑  作者: Mango

数组平衡索引的C程序

数组的平衡索引是这样一个索引,使得较低索引处的元素之和等于较高索引处的元素之和。例如,在数组 A 中:

例子 :

写一个函数int balance(int[] arr, int n) ;给定大小为 n 的序列 arr[],返回平衡索引(如果有),如果不存在平衡索引,则返回 -1。

方法一(简单但低效)
使用两个循环。外循环遍历所有元素,内循环确定外循环选取的当前索引是否为平衡索引。该解决方案的时间复杂度为 O(n^2)。

C
// C program to find equilibrium
// index of an array
  
#include 
  
int equilibrium(int arr[], int n)
{
    int i, j;
    int leftsum, rightsum;
  
    /* Check for indexes one by one until 
      an equilibrium index is found */
    for (i = 0; i < n; ++i) {       
  
        /* get left sum */
        leftsum = 0; 
        for (j = 0; j < i; j++)
            leftsum += arr[j];
  
        /* get right sum */
        rightsum = 0; 
        for (j = i + 1; j < n; j++)
            rightsum += arr[j];
  
        /* if leftsum and rightsum are same, 
           then we are done */
        if (leftsum == rightsum)
            return i;
    }
  
    /* return -1 if no equilibrium index is found */
    return -1;
}
  
// Driver code
int main()
{
    int arr[] = { -7, 1, 5, 2, -4, 3, 0 };
    int arr_size = sizeof(arr) / sizeof(arr[0]);
    printf("%d", equilibrium(arr, arr_size));
  
    getchar();
    return 0;
}


C
// C program to find equilibrium
// index of an array
  
#include 
  
int equilibrium(int arr[], int n)
{
    int sum = 0; // initialize sum of whole array
    int leftsum = 0; // initialize leftsum
  
    /* Find sum of the whole array */
    for (int i = 0; i < n; ++i)
        sum += arr[i];
  
    for (int i = 0; i < n; ++i) {
        sum -= arr[i]; // sum is now right sum for index i
  
        if (leftsum == sum)
            return i;
  
        leftsum += arr[i];
    }
  
    /* If no equilibrium index found, then return 0 */
    return -1;
}
  
// Driver code
int main()
{
    int arr[] = { -7, 1, 5, 2, -4, 3, 0 };
    int arr_size = sizeof(arr) / sizeof(arr[0]);
    printf("First equilibrium index is %d", 
                 equilibrium(arr, arr_size));
  
    getchar();
    return 0;
}


输出
3

时间复杂度: O(n^2)

方法2(棘手且高效)
这个想法是首先获得数组的总和。然后遍历数组并不断更新初始化为零的左和。在循环中,我们可以通过将元素一一相减得到正确的和。感谢 Sambasiva 提出这个解决方案并为此提供代码。

1) Initialize leftsum  as 0
2) Get the total sum of the array as sum
3) Iterate through the array and for each index i, do following.
    a)  Update sum to get the right sum.  
           sum = sum - arr[i] 
       // sum is now right sum
    b) If leftsum is equal to sum, then return current index. 
       // update leftsum for next iteration.
    c) leftsum = leftsum + arr[i]
4) return -1 
// If we come out of loop without returning then
// there is no equilibrium index

下图显示了上述方法的试运行:

下面是上述方法的实现:

C

// C program to find equilibrium
// index of an array
  
#include 
  
int equilibrium(int arr[], int n)
{
    int sum = 0; // initialize sum of whole array
    int leftsum = 0; // initialize leftsum
  
    /* Find sum of the whole array */
    for (int i = 0; i < n; ++i)
        sum += arr[i];
  
    for (int i = 0; i < n; ++i) {
        sum -= arr[i]; // sum is now right sum for index i
  
        if (leftsum == sum)
            return i;
  
        leftsum += arr[i];
    }
  
    /* If no equilibrium index found, then return 0 */
    return -1;
}
  
// Driver code
int main()
{
    int arr[] = { -7, 1, 5, 2, -4, 3, 0 };
    int arr_size = sizeof(arr) / sizeof(arr[0]);
    printf("First equilibrium index is %d", 
                 equilibrium(arr, arr_size));
  
    getchar();
    return 0;
}
输出
First equilibrium index is 3

输出:
第一个均衡指数是 3

时间复杂度: O(n)

有关详细信息,请参阅有关数组的 Equilibrium index 的完整文章!