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 :
Input: A[] = {-7, 1, 5, 2, -4, 3, 0}
Output: 3
3 is an equilibrium index, because:
A[0] + A[1] + A[2] = A[4] + A[5] + A[6]
Input: A[] = {1, 2, 3}
Output: -1
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!