求2個單鏈表的交集和并集

該程式生成 2 個鏈表的并集和交集。但是，它不會生成預期的輸出，其中不應存在串列 1 和串列 2 中的重復值，甚至不會顯示交集串列。我該如何解決這個問題？

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

typedef struct node{
    int data;
    struct node *next;
} node;

void insertAtBeg(node **head, int ele){
    node *newnode = (node*)malloc(sizeof(node));
    newnode->data = ele;
    newnode->next = (*head);
    (*head) = newnode;
}

int isPresent(node *temp, int ele){
    node *t = temp;
    while(t != NULL){
        if(t->data == ele)
            return 1;
        t = t->next;
    }
    return 0;
}

void printList(node *n){
    while(n != NULL){
        printf("%d->",n->data);
        n = n->next;
    }
}

node* getUnion(node *head1,node *head2){
    node *result = NULL;
    node *t1 = head1;
    node *t2 = head2;
    while(t1 != NULL){
        insertAtBeg(&result, t1->data);
        t1 = t1->next;
    }

    while(t2!=NULL){
        if(!isPresent(result, t2->data));
            insertAtBeg(&result,t2->data);
        t2 = t2->next;
    }
    return result;
}

node *getIntersection(node *head1,node *head2){
    node* result = NULL;
    node* t1 = head1;
    while(t1 != NULL){
        if(isPresent(head2, t1->data))
            insertAtBeg(&result,t1->data);
        t1 = t1->next;
    }
    return result;
}

int main(){
    node *intersection = NULL;
    node *unin = NULL;
    node *List1;
    node *List2;
    List1 = List2 = NULL;
    int i,n,m,temp;
    
    printf("Enter the size of the first linked list:\n");
    scanf("%d",&n);
    printf("Enter %d elements\n",n);
    for(i = 0;i < n;i  ){
        scanf("%d",&temp);
        insertAtBeg(&List1,temp);
    }
     printf("Displaying list 1:\n");
     printList(List1);

    printf("\nEnter the size of the second linked list:\n");
    scanf("%d",&m);
    printf("Enter %d elements\n",m);
    for(i = 0;i < m;i  ){
        scanf("%d",&temp);
        insertAtBeg(&List2,temp);
    }
    printf("Displaying list 2:\n");
    printList(List2);

    unin = getUnion(List1,List2);
    intersection = getIntersection(List1,List2);

    printf("\nLinked List with Union of List1 and List2:\n");
    printList(unin);
    printf("\nLinked List with Intersection of List1 and List2:\n");
    printList(intersection);
    return 0;
}

`

生成的輸出： 輸入第一個鏈表的大小：4 輸入4個元素 3 4 6 1 顯示串列1：1->6->4->3-> 輸入第二個鏈表的大小：4 輸入4個元素8 9 10 1 顯示串列 2：1->10->9->8-> List1 和 List2 并集的鏈表：8->9->10->1->3->4->6->1 -> 具有 List1 和 List2 交集的鏈表：1->

uj5u.com熱心網友回復：

函式內getUnion有錯別字

while(t2!=NULL){
    if(!isPresent(result, t2->data));
                                   ^^^^
        insertAtBeg(&result,t2->data);
    t2 = t2->next;
}

洗掉分號。

同樣在這個電話之后

printList(intersection);  

輸出換行符，例如

putchar( '\n' );

請注意，const如果串列未在函式內更改，例如，您應該宣告引數，這些引數表示指向帶有限定符的節點的指標

int isPresent( const node *temp, int ele){
    const node *t = temp;
    while(t != NULL){
        if(t->data == ele)
            return 1;
        t = t->next;
    }
    return 0;
}

void printList(const node *n){
    while(n != NULL){
        printf("%d->",n->data);
        n = n->next;
    }
}

node* getUnion( const node *head1,const node *head2){
    node *result = NULL;
    const node *t1 = head1;
    const node *t2 = head2;
    while(t1 != NULL){
        insertAtBeg(&result, t1->data);
        t1 = t1->next;
    }

    while(t2!=NULL){
        if(!isPresent(result, t2->data))
            insertAtBeg(&result,t2->data);
        t2 = t2->next;
    }
    return result;
}

node *getIntersection(const node *head1,const node *head2){
    node* result = NULL;
    const node* t1 = head1;
    while(t1 != NULL){
        if(isPresent(head2, t1->data))
            insertAtBeg(&result,t1->data);
        t1 = t1->next;
    }
    return result;
}

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