鏈表代碼不是非常優雅

Question

這是代碼的玩具版我真的想寫：

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

typedef struct node_s { 
    char val; 
    struct node_s *next; 
} node; 

int n; 

char obtain_next_char(){ 
    if (n<100) { 
     return 'a'; 
    } else { 
     return 0; 
    } 
} 

int main() { 

    char c; 
    node *start, *curr; 

    n = 0; 
    start = malloc(sizeof(node)); 
    curr = start; 

    while ((c=obtain_next_char()) != 0) { 
     curr->val = c; 
     curr->next = malloc(sizeof(node)); 
     curr = curr->next; 
    } 
    curr->next = NULL; 

    printf("chain made, %d links\n", n); 

    curr = start; 

    while (curr != NULL) { 
     printf(">>>%c<<<\n",curr->val); 
     curr=curr->next; 
    } 

} 

首先，關於obtain_next_char，該功能是非常簡化在這裏。有了真正的功能，就沒有辦法事先知道返回值是什麼。

現在，代碼可以工作，但它會在列表末尾產生一個空節點，我將不得不擺脫它。可以重新編寫這段代碼來彌補這個缺陷嗎？我試圖避免循環頭部的重複代碼，即我不想在循環外部設置第一個具有代碼的節點，然後在循環內部使用相同的代碼設置以下節點。

Answer 1

node dummy, *start, *curr; 

curr = &dummy; 

while ((c=obtain_next_char()) != 0) { 
    curr->next = malloc(sizeof(node)); 
    curr = curr->next; 
    curr->val = c; 
} 
curr->next = NULL; 
start = dummy.next; 

Answer 2

如何：

start = NULL; 
while ((c=obtain_next_char()) != 0) { 
    if (!start) { 
     start = malloc(sizeof(node)); 
     curr = start; 
    } else { 
     curr->next = malloc(sizeof(node)); 
     curr = curr->next; 
    } 
    curr->val = c; 
    curr->next = NULL; 
} 

Answer 3

我不知道你想要什麼來實現這個功能，因爲在這裏你不會在任何輸入掃描：

char obtain_next_char(){ 
    if (n<100) { 
     return 'a'; 
    } else { 
     return 0; 
    } 
} 

也許你需要這個：

char obtain_next_char() 
{ 
    char c; 

    scanf(" %c",&c); //scanning input 

    if (n<100) //if number of links < 100 then, return character 
    { 
     return c; 
    } 
    else //else return '0' (ascii character) 
    { 
     return '0'; 
    } 
} 

---

，但它在列表的最後導致一個空節點

原因：

的問題是，被在年底創建了一個新的節點，而循環

while ((c=obtain_next_char()) != 0) 
{ 
    curr->val = c; 
    curr->next = malloc(sizeof(node)); //new node at end of each iteration 
    curr = curr->next; 
} 

所以，當你結束輸入，最後一個空節點仍然

---

溶液：（不與代碼的循環外第一節點）

爲了避免這個問題，嘗試在每個迭代

while ((c=obtain_next_char()) != '0') 
{ 
     if(n != 0) //to avoid again creating for head node 
     { 
      curr->next = malloc(sizeof(node)); 
      curr = curr->next; 
     } 
     curr->val = c; 
     n++; //increasing number of links 
} 

• 這種方式，開始創建節點，您可以避免在每次迭代結束時創建額外的空節點。

---

我不想建立與代碼的第一個節點的循環

while ((c=obtain_next_char()) != '0') 
{ 
     if(n != 0) //any other node 
     { 
      curr->next = malloc(sizeof(node)); 
      curr = curr->next; 
     } 
     else //for head node 
     { 
      start = malloc(sizeof(node)); 
      curr = start; 
     } 
     curr->val = c; 
     n++; //increasing number of links made 
} 

---

乾脆把你的外碼將是：

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

typedef struct node_s { 
    char val; 
    struct node_s *next; 
} node; 

int n; 

char obtain_next_char() 
{ 
    char c; 

    scanf(" %c",&c); 

    if (n<100) 
    { 
     return c; 
    } 
    else 
    { 
     return '0'; 
    } 
} 

int main() { 

    char c; 
    node *start=NULL, *curr=NULL; 

    n = 0; 


    while ((c=obtain_next_char()) != '0') 
    { 
     if(n != 0) //any other node 
     { 
      curr->next = malloc(sizeof(node)); 
      curr = curr->next; 
     } 
     else //for head node 
     { 
      start = malloc(sizeof(node)); 
      curr = start; 
     } 

     curr->val = c; 
     n++; 
    } 

    curr->next = NULL; 

    printf("chain made, %d links\n", n); 

    curr = start; 

    while (curr != NULL) { 
     printf(">>>%c<<<\n",curr->val); 
     curr=curr->next; 
    } 

} 

輸入：

1 2 3 4 5 0 

輸出：

chain made, 5 links 
>>>1<<< 
>>>2<<< 
>>>3<<< 
>>>4<<< 
>>>5<<<