单链表删除一个节点

扩展问题。

1.单链表从尾到头输出

解法1.用下面的rev，然后正向输出链表

解法2.用栈或者递归正向输出链表

比如用递归：

void fun(link* p){

　　if(p->next!=null){

　　　　fun(p->next);

　　}

　　printf("%d\t",p->value);

}

 

2.反向输出字符串

可以参照解法2.

void fun(char* s){

　　if(*s != '\0'){

　　　　fun(s+1);

　　}

　　printf("%c",*s);

}

3.给定待删除节点的指针，在o(1)时间删除节点。

　

p如果不是最后一个节点，就把p当作前驱，把p->next的数据域复制过来，然后就和普通删除一样了。p要是最后一个节点，就只能遍历找前驱了。

　

有头结点的情况，附加一个逆置

#include <stdio.h>
#include <malloc.h>
#define null 0
typedef struct Node{
    int value;
    struct Node* next;
};

//尾插 
void create(struct Node* head){
    //p是工作指针 
    struct Node* p = head;
    int i;
    for(i = 1; i<=1; i++){
    
        p->next = malloc(sizeof(struct Node));
        p->next->value = i;
        p->next->next = null;
        p = p->next;            
        
    }    
}

void print(struct Node* head){
    struct Node* p = head->next;
    while(p){
        printf("%d\t",p->value);
        p = p->next;
    }
    printf("\n");
    
}

int delete(struct Node* x,struct Node* head){
    struct Node* p = head->next;
    struct Node* pre = head;
    
    while(p){
        if(p->value == x->value){
            pre->next = p->next;
            free(p);
            return 1;
        }else{
            pre = p;
        }
        
        p = p->next;
    }
    
    return 0;
}

void rev(struct Node* head){
    if(head->next != null){
        struct Node* p = head->next->next;
        struct Node* temp;
        struct Node* pre = head->next;
        pre->next = null;
        while(p){
            temp = p;
            p = p->next;
            temp->next = pre;
            pre = temp;
        }
        head->next = pre;    
    }    
}

int main(int argc,char *argv[]){
    struct Node* head = malloc(sizeof(struct Node));
    head->value = -1;
    create(head);
    print(head);
    
    
    struct Node* x = malloc(sizeof(struct Node));
    x->value = 1;
    delete(x,head);
    print(head);
    
    
    rev(head);
    print(head);
    
    return 0;    
}

没有头结点的情况，还是有些不同的地方要注意的。

#include <stdio.h>
#include <malloc.h>
#define null 0
struct Node{
    int value;
    struct Node* next;
};

//头插 
struct Node* create(){
    struct Node* head = null;
    int i;
    for(i = 1; i<=10; i++){
        if(head == null){
            head = malloc(sizeof(struct Node));
            head->value = i;
            head->next = null;    
        }else{
            struct Node* p = malloc(sizeof(struct Node));    
            p->value = i;
            p->next    = head->next;                
            head->next = p;
        }
    }
    return head;    
}

void print(struct Node* head){
    struct Node* p = head;
    while(p){
        printf("%d\t",p->value);
        p = p->next;
    }
    printf("\n");
    
}

//注意这里想改变head指针的指向 必须传head的引用 所以是 Node** 
int delete(struct Node* x,struct Node** head){
    struct Node* p = head;
    struct Node* pre = head;
    //被删除节点是第一个的时候要特殊处理 
    if((*head)->value == x->value    ){
            *head = (*head)->next;
            free(p);
            return 1;
    }
    while(p){
        if(p->value == x->value){
            pre->next = p->next;
            free(p);
            return 1;    
        }else{
            pre = p;
        }
        
        p = p->next;
    }
    
    return 0;
}

int main(int argc,char *argv[]){
    
    struct Node* head = create();
    print(head);
    
    struct Node* x = malloc(sizeof(struct Node));
    x->value = 1;
    delete(x,&head);
    print(head);
    
    return 0;    
}