單鏈表反轉
實現(xiàn)代碼
#include <stdio.h>
#include <stdlib.h>
struct node {
int data;
struct node *next;
};
struct node* list_init(int length)
{
int i = 0;
struct node *p, *q;
if(length <= 0) {
return NULL;
}else{
p = malloc(sizeof(struct node));
p->data = 0;
}
for(i=1, q=p; i<length; i++) {
q->next = malloc(sizeof(struct node));
q = q->next;
q->data = i;
}
q->next = NULL;
return p;
}
int list_print(struct node *pnode)
{
int i;
for(i=0; pnode; i++) {
printf("node[%i]: %i\n", i, pnode->data);
pnode = pnode->next;
}
return 0;
}
struct node* list_reverse(struct node *pnode)
{
struct node *p, *q = NULL;
while( pnode ) {
p = pnode->next;
pnode->next = q;
q = pnode;
pnode = p;
}
return q;
}
int list_free(struct node *pnode)
{
struct node *p, *q;
p = pnode;
while(p) {
q = p->next;
free(p);
p = q;
}
return 0;
}
int main(int argc, char* argv[])
{
int len;
struct node *pnode = NULL;
printf("input number:");
scanf("%i", &len);
pnode = list_init(len);
list_print(pnode);
list_print(list_reverse(pnode));
list_free(pnode);
}
尚未實現(xiàn)
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