针对考研的C语言学习（2019链表大题）

题目解析：

【考】双指针算法，逆置法，归并法。

解析：因为题目要求空间复杂度为O(1)，即不能再开辟一条链表，因此我们只能用变量来整体挪动原链表。

第一步先找出中间节点

typedef NODE* Node;
Node find_mid(Node& h)
{
Node pre, cur;
pre = h->next;
cur = pre;
while (cur)
{
cur = cur->next;
if (!cur) break;
cur = cur->next;
if (!cur) break;
pre = pre->next;
}
Node l2 = (Node)malloc(sizeof(NODE));
l2->next = pre->next;
pre->next = NULL;
return l2;
}

第二步：把找出的中间节点之后的组成的新链表原地逆置

void reverse_list(Node& l)
{
Node s, r, t;
s = l->next, r = s->next, t = r->next;
s->next = NULL;
while (t)
{
r->next = s;
s = r;
r = t;
t = t->next;
}
r->next = s;
l->next = r;
}

第三步：合并链表

void merge_list(Node& l1, Node& l2)
{
Node a =  l1->next, b = l2->next;
Node acur = a;
a = a->next;
while (a && b)
{
acur->next = b;
b = b->next;
acur = acur->next;
acur->next = a;
a = a->next;
acur = acur->next;
}
if(!a) acur->next = b;

}

【注】以上三步核心算法即为笔试时写的答案

为了让读者看清正确性，我写出完整能运行的代码供参考

#define _CRT_SECURE_NO_WARNINGS 1
#include<stdio.h>
#include<stdlib.h>
typedef int ElemType;
typedef struct LNode {
ElemType data;
struct LNode* next;
}NODE;
typedef NODE* Node;

void insert_head_list(Node& l)
{
ElemType data = 0;
scanf("%d", &data);
while (data != 9999)
{
Node tmp = (Node)malloc(sizeof(NODE));
tmp->data = data;
tmp->next = l->next;
l->next = tmp;
scanf("%d", &data);
}
}


void print_list(Node& l)
{
Node cur = l->next;
while (cur)
{
printf("%d ", cur->data);
cur = cur->next;
}
printf("\n");
}


Node find_mid(Node& h)
{
Node pre, cur;
pre = h->next;
cur = pre;
while (cur)
{
cur = cur->next;
if (!cur) break;
cur = cur->next;
if (!cur) break;
pre = pre->next;
}
Node l2 = (Node)malloc(sizeof(NODE));
l2->next = pre->next;
pre->next = NULL;
return l2;
}
void reverse_list(Node& l)
{
Node s, r, t;
s = l->next, r = s->next, t = r->next;
s->next = NULL;
while (t)
{
r->next = s;
s = r;
r = t;
t = t->next;
}
r->next = s;
l->next = r;
}

void merge_list(Node& l1, Node& l2)
{
Node a =  l1->next, b = l2->next;
Node acur = a;
a = a->next;
while (a && b)
{
acur->next = b;
b = b->next;
acur = acur->next;
acur->next = a;
a = a->next;
acur = acur->next;
}
if(!a) acur->next = b;

}
int main()
{
Node l = (Node)malloc(sizeof(NODE)); //存储头节点的头指针
l->next = NULL;
insert_head_list(l);//头插法
print_list(l);
Node l2 = find_mid(l);
/*print_list(l);*/
print_list(l2);
reverse_list(l2);
print_list(l2);
merge_list(l, l2);
print_list(l);
return 0;
}

运行结果图片

链表长度为奇数时

链表长度为偶数时

原文地址：https://blog.csdn.net/Loyalty_lu/article/details/142670020

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