//删除带头结点的链表中的结点
note:这里需要三个指针,用递归法只需一个*q,遍历用L->next即可。
#include<iostream>
using namespace std;
typedef struct lnode {
int data;
struct lnode* next;
}lnode,*linklist;
int a[5] = { 1,2,1,1,3 };
int n = 5;
void buildlist(linklist& L) {
L = (lnode*)malloc(sizeof(lnode));
if (L == NULL)return;//判空
lnode* r = L, * s;
for (int i = 0; i < n; i++) {
s = (lnode*)malloc(sizeof(lnode));
s->data = a[i];
r->next = s;
r = r->next;
}
r->next = NULL;
}
void disp(linklist L) {
lnode* s = L->next;
while (s) {
cout << s->data << " ";
s = s->next;
}
cout << endl;
}
void deletex(linklist& L, int x) {
if (L == NULL)return;//判空
lnode* q, * r = L, * p = L->next;//q用来释放空间,r表示要删除元素x的前序,p用来遍历链表
while (p) {
if (p->data != x) {
r->next = p;//当删除元素x时,把x的前元素跟x的后元素相连
r = p;
p = p->next;
}
else {
q = p;
p = p->next;
free(q);
}
}
}
int main()
{
linklist L;
buildlist(L);
disp(L);
deletex(L, 1);
disp(L);
return 0;
}
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