在我的 c 代码中删除重复项时出错

Question

所以我的代码几乎是完整的，在用于搜索重复项的 for 循环中，我的代码有一个小的逻辑错误。 我可以得到任何关于我出错的提示吗？

#include <stdio.h>
#include <stdlib.h>

struct studentID{
    int value;          //a data member which is an integer
    struct studentID *next;         //a data member which is a pointer to next node
};

typedef struct studentID STUDENTID;     //creating a nickname for struct studentID as STUDENTID
typedef STUDENTID *STUDENTIDPtr;        //creating a nickname for STUDENTID as STUDENTIDPtr

//Global variables
STUDENTIDPtr previousPtr;           //pointer to previous node in list
STUDENTIDPtr currentPtr;            //pointer to current node in list

int main(){
    int checker[5];
    int removeDuplicate[5];

    STUDENTIDPtr newPtr1;           //creating a pointer to create a new node
    STUDENTIDPtr newPtr2;           //creating a pointer to create a new node
    STUDENTIDPtr newPtr3;           //creating a pointer to create a new node
    STUDENTIDPtr newPtr4;           //creating a pointer to create a new node
    STUDENTIDPtr newPtr5;           //creating a pointer to create a new node

    //creation of the first node
    newPtr1 = malloc(sizeof(STUDENTID));            //This is when a node1 is created
    newPtr2 = malloc(sizeof(STUDENTID));            //This is when a node2 is created
    newPtr3 = malloc(sizeof(STUDENTID));            //This is when a node3 is created
    newPtr4 = malloc(sizeof(STUDENTID));            //This is when a node4 is created
    newPtr5 = malloc(sizeof(STUDENTID));            //This is when a node5 is created

    newPtr1 -> value = 5; // assign data in node 01
    newPtr1 -> next = newPtr2;

    newPtr2 -> value = 2; // assign data in node 02 
    newPtr2 -> next = newPtr3;

    newPtr3 -> value = 4; // assign data in node 03 
    newPtr3 -> next = newPtr4;

    newPtr4 -> value = 4; // assign data in node 04
    newPtr4 -> next = newPtr5;

    newPtr5 -> value = 1; // assign data in node 05
    newPtr5 -> next = NULL;

    currentPtr = newPtr1;
    printf("n");
    printf("n");
    /*Loop to print the last 5 digits in my student ID*/
    printf("The last 05 digits of my Student ID are,n");
    while (currentPtr != NULL){         //while not the end of the list
        printf("%d - ", currentPtr->value);
        currentPtr = currentPtr ->next;
    }
    printf("n");
    printf("n");
    /*Assigning the 5 digits into an array*/
    checker[0] = newPtr1 -> value;
    checker[1] = newPtr2 -> value;
    checker[2] = newPtr3 -> value;
    checker[3] = newPtr4 -> value;
    checker[4] = newPtr5 -> value;

    /*This is a loop just to show that the values have successfully been addded to the array */
    printf("The List before removing duplicate digits is: ");
    for(int i=0; i<5; i++){
        printf("%d", checker[i]);
    }

    /*This would be the loop to remove duplicates*/
    for(int i=0; i<=5; i++){
        int temp;
        temp = checker[i];
        for(int j=0; j<=5; j++){
            if(temp != checker[j]){
                removeDuplicate[j] = temp;
            }
        }
    }
    printf("n");
    printf("n");
    /*This would be the loop to print the finalized list without duplicates*/
    printf("The List after removing duplicate digits is: ");
    for(int i=0; i<5; i++){
        printf("%d", removeDuplicate[i]);
    }
    printf("n");
    return 0;
}

现在一切似乎都很好，但我确信我的 for 循环搜索重复项是我出错的地方。我的方法是取一位数字并将其与其他数字进行比较。 如果它没有任何重复项，则将其放入数组中并移动到下一个数字。

作为最终列表，它必须仅包含 4 位数字作为The List after removing duplicate digits is: 5241

Answer 1

对于初学者来说，数组与列表没有任何共同点，除了它们存储数据成员的值value列表节点的值。因此，您不会删除列表中的重复项。

数组removeDuplicate未初始化

int removeDuplicate[5];

所以这个循环

    for(int j=0; j<=5; j++){
        if(temp != checker[j]){
            removeDuplicate[j] = temp;
        }
    }

具有未定义的行为。此外，循环j<=5中的此条件提供了超出阵列的访问。

循环可能如下所示

int n = 0;
for ( int i = 0; i < 5; i++ )
{
    int j = 0;
    while ( j < n && checker[i] != removeDuplicate[j] ) j++; 
    if ( j == n ) removeDuplicate[n++] = checker[i];
}

然后

printf("The List after removing duplicate digits is: ");
for( int i = 0; i < n; i++){
    printf("%d", removeDuplicate[i]);
}