我正在尝试编写一个递归函数,该函数检查给定数组是否在偶数索引上向上排序。例如,我有一个size 5
数组。数字为1 , 2 , 3 , 4 ,5
。它将return 1
,因为arr[0] < arr[2] < arr[4]
。
该行的问题:
if (i >= arr[isEven]) return 1; //Sorted
以下是代码:
#include <stdio.h>
#include <conio.h>
int SortedUpDown(int arr, int num);
int main()
{
int sizeOfArr = 0, i = 0,*arr, num = 0, checkIfSorted = -1;
printf("Enter the size of the array : \n");
scanf("%d", &sizeOfArr);
printf("Enter %d numbers to the array \n: ", sizeOfArr);
arr = (int *)malloc(sizeOfArr * sizeof(int));
for (i = 0; i < sizeOfArr; i++)
{
scanf("%d", &num);
*(arr + i) = num;
printf("%d ", arr[i]);
}
checkIfSorted = SortedUpDown(*arr,sizeOfArr);
getch();
return 0;
}
int SortedUpDown(int *arr, int num)
{
int i = 0, isEven = 0;
if (num % 2 == 0) isEven = num - 2; //if The array is even, (for example arr[6] then check 0,1,2,3,4,5 only 0,2,4)
else isEven = num - 1; //if the array is odd, (for example arr[7] then check 0,1,2,3,4,5,6 only 0,2,4,6)
if (i >= arr[isEven]) return 1; //Sorted
if (arr[i] > arr[i + 2]) return 0; //Not sorted
return i += 2, SortedUpDown(arr, num); //Advance i by 2.
}
抛出错误:
Exception thrown at 0x00AE18BC in ConsoleApplication1.exe: 0xC0000005: Access violation reading location 0x00000011.
If there is a handler for this exception, the program may be safely continued.
编辑:我想现在检查奇数索引是否向下并且它不起作用,这是我的代码:
我有两个问题:
为什么它不起作用?
#include <stdio.h> #include <stdlib.h> #include <conio.h> int SortedDown(int *arr, int num); int SortedUp(int *arr, int num); int main(void) { int sizeOfArr = 0, i = 0, *arr, num = 0, checkIfSortedDown = -1, checkIfSortedUp = -1; printf("Enter the size of the array : \n"); scanf("%d", &sizeOfArr); printf("Enter %d numbers to the array \n: ", sizeOfArr); arr = (int *)malloc(sizeOfArr * sizeof(int)); for (i = 0; i < sizeOfArr; i++) { scanf("%d", arr + i); printf("%d ", arr[i]); } puts(""); checkIfSortedDown = SortedDown(arr, sizeOfArr); checkIfSortedUp = SortedUp(arr, sizeOfArr); if (checkIfSortedDown && checkIfSortedUp) puts("Sorted"); else puts("Not Sorted"); free(arr); getch(); return 0; } int SortedDown(int *arr, int num) { --num;//size to last index if (num % 2 != 0) //if index is not even --num; if (num <= 0) return 1;//Sorted else if (arr[num - 2] > arr[num]) return 0;//Not sorted else return SortedDown(arr, num - 2 + 1);//+1 : last index to size } int SortedUp(int *arr, int num) { --num;//size to last index if (num % 2 == 0) --num; if (num <= 0) return 1;//Sorted else if (arr[num - 2] < arr[num]) return 0;//Not sorted else return SortedUp(arr, num - 2 + 1);//+1 : last index to size }
答案 0 :(得分:2)
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答案 1 :(得分:1)
当您从SortedUpDown()中调用SortedUpDown()时,看起来您遇到了无限循环,因为在每次调用时我都设置为零。
尝试使用static来保留'i'的值;
static int i = 0;
编辑:
此外,我修复了一些问题,从fn声明开始作为指针。
int SortedUpDown(int *arr, int num);
并称之为:
checkIfSorted = SortedUpDown(arr,sizeOfArr);
以下是完整代码:
#include <stdio.h>
#include <stdlib.h>
int SortedUpDown(int *arr, int num);
int main()
{
int sizeOfArr = 0, i = 0,*arr, num = 0, checkIfSorted = -1;
printf("Enter the size of the array : \n");
scanf("%d", &sizeOfArr);
printf("Enter %d numbers to the array \n: ", sizeOfArr);
arr = (int *)malloc(sizeOfArr * sizeof(int));
for (i = 0; i < sizeOfArr; i++)
{
scanf("%d", &num);
*(arr + i) = num;
printf("%d ", arr[i]);
}
checkIfSorted = SortedUpDown(arr,sizeOfArr);
printf("\ndone %d\n",checkIfSorted);
//getch();
return 0;
}
int SortedUpDown(int *arr, int num)
{
static int i = 0;
if (i >= num - 1) return 1;
if (arr[i] > arr[i + 1]) return 0;
return i++, SortedUpDown(arr, num);
}