现有数组a[8]={1,3,8,9,6,2,4,7},请使用编写顺序查找算法实现查找数字2的位置。
时间: 2024-04-07 08:24:46 浏览: 26
以下是使用顺序查找算法实现查找数字2的位置的代码:
```python
def sequential_search(arr, target):
for i in range(len(arr)):
if arr[i] == target:
return i
return -1
a = [1, 3, 8, 9, 6, 2, 4, 7]
target =
position = sequential_search(a, target)
print("The position of", target, "in the array is:", position)
```
输出结果为:The position of 2 in the array is: 5
相关问题
现有数组a[8]={1,3,8,9,6,2,4,7},请使用编写折半算法查找6的位置。
以下是使用折半算法查找数组中元素位置的示例代码:
```c
#include <stdio.h>
int binarySearch(int arr[], int left, int right, int target) {
while (left <= right) {
int mid = left + (right - left) / 2;
if (arr[mid] == target) {
return mid;
}
if (arr[mid] < target) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return -1;
}
int main() {
int arr[] = {1, 3, 8, 9, 6, 2, 4, 7};
int target = 6;
int size = sizeof(arr) / sizeof(arr[0]);
int position = binarySearch(arr, 0, size - 1, target);
if (position != -1) {
printf("The position of %d is %d\n", target, position);
} else {
printf("%d is not found in the array\n", target);
}
return 0;
}
```
现有数组a[8]={1,3,8,9,6,2,4,7},请使用编写冒泡排序算法实现从小到大排序。
以下是使用冒泡排序算法实现从小到大排序的代码:
```python
a = [1, 3, 8, 9, 6, 2, 4, 7]
for i in range(len(a)-1):
for j in range(len(a)-1-i):
if a[j] > a[j+1]:
a[j], a[j+1] = a[j+1], a[j]
print("排序后的数组:", a)
```
输出结果为:[1, 2, 3, 4, 6, 7, 8, 9]