C语言强化——链表（1）
目录
- 链表的增删（不带头结点）
- 链表相关面试题
- 合并两个有序链表
- 单链表原地逆置
- 找出链表的倒数第四个节点
- 找出链表的中间节点
- 判断单链表是否有环
- 求链表交点
- 删除有序单链表中重复的元素
- 链表按奇数、偶数值拆分
链表的增删（不带头结点）
func.h
#include<stdio.h>
#include<string.h>
typedef struct node {
int val;
struct node * next;
}Node, *pNode;
void listTailInsert(pNode*, pNode*, int);
void listHeadInsert(pNode*, pNode*, int);
void listSortInsert(pNode*, pNode*, int);
void listDeleteNode(pNode*, pNode*, int);
void listPrint(pNode head);
func.c
#include "func.h"
void listTailInsert(pNode *head, pNode *tail, int val)
{
pNode newNode = (pNode)calloc(1, sizeof(Node));
newNode->val = val;
if (NULL == *tail) { //链表为空
*head = newNode;
*tail = newNode;
}
else {
(*tail)->next = newNode;
*tail = newNode;
}
}
void listHeadInsert(pNode *head, pNode *tail, int val)
{
pNode newNode = (pNode)calloc(1, sizeof(Node));
newNode->val = val;
if (NULL == *head) {
*head = newNode;
*tail = newNode;
}
else {
newNode->next = *head;
*head = newNode;
}
}
void listSortInsert(pNode *head, pNode *tail, int val)
{
pNode newNode = (pNode)calloc(1, sizeof(Node));
newNode->val = val;
pNode pCur, pPre;
pCur = pPre = *head;
if (NULL == *head) {
*head = newNode;
*tail = newNode;
}
else if (pCur->val > val) { //头插
newNode->next = *head;
*head = newNode;
}
else {
while (pCur) { //中间插入
if (val < pCur->val) {
pPre->next = newNode;
newNode->next = pCur;
break;
}
pPre = pCur;
pCur = pCur->next;
}
if (NULL == pCur) { //尾部插入
pPre->next = newNode;
*tail = newNode;
}
}
}
void listDeleteNode(pNode *head, pNode *tail, int val)
{
pNode pPre, pCur;
pPre = pCur = *head;
if (pCur == NULL) {
printf("list is empty!\n");
//printf("Delete failed , doesn't find the node!\n");
return;
}
else if (pCur->val == val) { //删除的是头部
*head = pCur->next;
}
else {
while (pCur) {
if (pCur->val == val) {
pPre->next = pCur->next;
break;
}
pPre = pCur;
pCur = pCur->next;
}
if (pCur == *tail) {
*tail = pPre;
}
if (NULL == pCur) {
printf("Don't find the node!\n");
}
}
}
void listPrint(pNode head)
{
while (head) {
printf("%3d", head->val);
head = head->next;
}
printf("\n");
}
main.c
#include "func.h"
int main() {
printf("please input the key of node:\n");
pNode head = NULL, tail = NULL;
int val;
while (scanf("%d", &val) != EOF) {
//listTailInsert(&head, &tail, val);
//listHeadInsert(&head, &tail, val);
listSortInsert(&head, &tail, val);
}
while (printf("please input delete num:\n"), scanf("%d", &val) != EOF){
listDeleteNode(&head, &tail, val);
listPrint(head);
}
listPrint(head);
return 0;
}
链表相关面试题
- 合并两个有序链表
#include<stdio.h>
#include<string.h>
typedef struct node {
int val;
struct node *next;
}Node, *pNode;
void listTailInsert(pNode *head, pNode *tail, int val)
{
pNode newNode = (pNode)calloc(1,sizeof(Node));
newNode->val = val;
if (NULL == *tail) { //链表为空
*head = newNode;
*tail = newNode;
}
else {
(*tail)->next = newNode;
*tail = newNode;
}
}
pNode head = NULL, tail = NULL;
void Union(pNode head1, pNode tail1, pNode head2, pNode tail2, int len1, int len2) {
int i = 0;
while (head1 && head2) {
if (head1->val >= head2->val) {
listTailInsert(&head, &tail, head2->val);
head2 = head2->next;
}
else {
listTailInsert(&head, &tail, head1->val);
head1 = head1->next;
}
}
while (head1) {
listTailInsert(&head, &tail, head1->val);
head1 = head1->next;
}
while (head2) {
listTailInsert(&head, &tail, head2->val);
head2 = head2->next;
}
}
void listPrint(pNode head)
{
while (head) {
printf("%3d", head->val);
head = head->next;
}
printf("\n");
}
int main() {
int a[5] = { 1,2,7,9,13 };
int b[8] = { 3,4,6,8,9,11,15,19 };
pNode head1, head2, tail1, tail2;
head1 = head2 = tail1 = tail2 = NULL;
for (int i = 0;i < 5;++i) {
listTailInsert(&head1, &tail1, a[i]);
}
for (int i = 0;i < 8;++i) {
listTailInsert(&head2, &tail2, b[i]);
}
printf("两个有序链表：\n");
listPrint(head1);
listPrint(head2);
printf("合并后：\n");
Union(head1, tail1, head2, tail2, 5, 8);
listPrint(head);
return 0;
}
- 单链表原地逆置
（三指针法：用三个指针分别指向前三个节点，记为A、B、C，第一步让B的next指向A，A的next置为NULL，接着用循环，每次让B的next指向A，并且让三个指针同时往后移一个为止，直到c为NULL为止。）
#include<stdio.h>
typedef struct node {
int val;
struct node *next;
}Node, *pNode;
void listTailInsert(pNode *head, pNode *tail, int val)
{
pNode newNode = (pNode)calloc(1, sizeof(Node));
newNode->val = val;
if (NULL == *tail) { //链表为空
*head = newNode;
*tail = newNode;
}
else {
(*tail)->next = newNode;
*tail = newNode;
}
}
void listPrint(pNode head)
{
while (head) {
printf("%3d", head->val);
head = head->next;
}
printf("\n");
}
pNode* reverse(pNode *head) {
pNode a = *head, b = (*head)->next, c = b->next;
b->next = a;
a->next = NULL;
while (c) {
a = b;
b = c;
c = c->next;
b->next = a;
}
return b;
}
int main() {
int a[5] = { 1,9,3,-2,7 };
pNode head, tail;
head = tail = NULL;
for (int i = 0;i < 5;i++) {
listTailInsert(&head, &tail, a[i]);
}
listPrint(head);
pNode t = reverse(&head);
listPrint(t);
return 0;
}
- 找出链表的倒数第四个节点
（双指针法，让第一个结点先移动4次，第二个结点再出发）
#include<stdio.h>
typedef struct node {
int val;
struct node *next;
}Node, *pNode;
void listTailInsert(pNode *head, pNode *tail, int val)
{
pNode newNode = (pNode)calloc(1, sizeof(Node));
newNode->val = val;
if (NULL == *tail) { //链表为空
*head = newNode;
*tail = newNode;
}
else {
(*tail)->next = newNode;
*tail = newNode;
}
}
void listPrint(pNode head)
{
while (head) {
printf("%3d", head->val);
head = head->next;
}
printf("\n");
}
int main() {
int a[10] = { 1,9,3,-2,7,99,6,-3,5,0 };
pNode head, tail, last;
head = tail = NULL;
for (int i = 0;i < 10;i++) {
listTailInsert(&head, &tail, a[i]);
}
listPrint(head);
last = head;
for (int i = 0;i < 4;i++) {
head = head->next;
}
while (head) {
head = head->next;
last = last->next;
}
printf("倒数第四个节点的值为%d\n", last->val);
return 0;
}
- 找出链表的中间节点
（双指针法：第一个指针每次移动2步，第二个指针每次移动1步）
#include<stdio.h>
#include<string.h>
typedef struct node {
int val;
struct node *next;
}Node, *pNode;
void listTailInsert(pNode *head, pNode *tail, int val)
{
pNode newNode = (pNode)calloc(1, sizeof(Node));
newNode->val = val;
if (NULL == *tail) { //链表为空
*head = newNode;
*tail = newNode;
}
else {
(*tail)->next = newNode;
*tail = newNode;
}
}
void listPrint(pNode head)
{
while (head) {
printf("%3d", head->val);
head = head->next;
}
printf("\n");
}
void findMidNode(pNode head) {
pNode t1, t2;
t1 = t2 = head;
while (t2) {
t2 = t2->next;
t2 = t2->next;
t1 = t1->next;
}
printf("中间节点为%d\n", t1->val);
}
int main() {
int a[11] = { 1,9,3,-2,7,99,6,-3,5,0,-8 };
pNode head, tail, last;
head = tail = NULL;
for (int i = 0;i < 10;i++) {
listTailInsert(&head, &tail, a[i]);
}
listPrint(head);
findMidNode(head);
return 0;
}
- 判断单链表是否有环
（双指针法，快指针每次移动2步，慢指针每次移动1步，若有环快指针必定会在一圈内与慢指针相遇）
#include<stdio.h>
#include<string.h>
typedef struct node {
int val;
struct node *next;
}Node, *pNode;
void listTailInsert(pNode *head, pNode *tail, int val)
{
pNode newNode = (pNode)calloc(1, sizeof(Node));
newNode->val = val;
if (NULL == *tail) { //链表为空
*head = newNode;
*tail = newNode;
}
else {
(*tail)->next = newNode;
*tail = newNode;
}
}
void listPrint(pNode head)
{
while (head) {
printf("%3d", head->val);
head = head->next;
}
printf("\n");
}
int judge(pNode head) {
pNode slow, fast;
slow = fast = head;
while (fast) {
slow = slow->next;
fast = fast->next;
fast = fast->next;
if (slow == fast) {
printf("有环！\n");
return 0;
}
}
printf("无环！\n");
return 1;
}
int main() {
int a[11] = { 1,9,3,-2,7,99,6,-3,5,0,-8 };
int b[11] = { 1,9,3,-2,7,99,6,-3,5,0,-8 };
pNode head, tail;
head = tail = NULL;
for (int i = 0;i < 10;i++) {
listTailInsert(&head, &tail, a[i]);
}
if (judge(head)) { //无环，输出链表
listPrint(head);
}
printf("------------------\n");
pNode head2, tail2;
head2 = tail2 = NULL;
for (int i = 0;i < 10;i++) {
listTailInsert(&head2, &tail2, b[i]);
}
tail2->next = head2; //构造环
if (judge(head2)) { //有环不输出
listPrint(head2);
}
return 0;
}
- 求两个链表的相交结点
（维护两个指针，先让长的链表的头指针往后移，直至和短链表的长度相同，再同时将两个链表的指针比较并后移）
#include<stdio.h>
#include<string.h>
typedef struct node {
int val;
struct node *next;
}Node, *pNode;
void listTailInsert(pNode *head, pNode *tail, int val)
{
pNode newNode = (pNode)calloc(1, sizeof(Node));
newNode->val = val;
if (NULL == *tail) { //链表为空
*head = newNode;
*tail = newNode;
}
else {
(*tail)->next = newNode;
*tail = newNode;
}
}
void listPrint(pNode head)
{
while (head) {
printf("%3d", head->val);
head = head->next;
}
printf("\n");
}
void findcommon(pNode head1, pNode head2, int len1, int len2) {
if (len1 > len2) {
while (len1 - len2) {
head1 = head1->next;
--len1;
}
}
else {
while (len2 - len1) {
head2 = head2->next;
--len2;
; }
}
while (head1&&head2) {
if (head1 == head2) {
printf("公共节点是%d\n", head1->val);
return;
}
head1 = head1->next;
head2 = head2->next;
}
printf("无公共节点！\n");
return;
}
int main() {
int a[5] = { 1,2,7,9,13 };
int b[10] = { 3,4,6,8,9,11,15,19,9,2 };
int c[5] = { 20,21,22,23,24 };
pNode head1, head2, tail1, tail2;
head1 = head2 = tail1 = tail2 = NULL;
for (int i = 0;i < 5;++i) {
listTailInsert(&head1, &tail1, a[i]);
}
for (int i = 0;i < 10;++i) {
listTailInsert(&head2, &tail2, b[i]);
}
findcommon(head1, head2, 5, 10);
pNode head3, tail3;
head3 = tail3 = NULL;
for (int i = 0;i < 5;++i) {
listTailInsert(&head3, &tail3, c[i]);
}
tail1->next = head3;
tail2->next = head3;
findcommon(head1, head2, 10, 15);
return 0;
}
- 删除有序单链表中的重复元素
（双指针）
#include<stdio.h>
#include<string.h>
typedef struct node {
int val;
struct node *next;
}Node, *pNode;
void listTailInsert(pNode *head, pNode *tail, int val)
{
pNode newNode = (pNode)calloc(1, sizeof(Node));
newNode->val = val;
if (NULL == *tail) {
*head = newNode;
*tail = newNode;
}
else {
(*tail)->next = newNode;
*tail = newNode;
}
}
void listPrint(pNode head)
{
while (head) {
printf("%3d", head->val);
head = head->next;
}
printf("\n");
}
void DeleteCommon(pNode *head) {
pNode pCur, pPre, pFree;
pPre = *head;
pCur = (*head)->next;
while (pCur) {
if (pCur->val == pPre->val) {
pFree = pCur;
pPre->next = pCur->next;
pCur = pCur->next;
free(pFree);
}
else {
pPre = pCur;
pCur = pCur->next;
}
}
}
int main() {
int a[10] = { 1,3,3,5,5,8,9,9,11,11 };
pNode head, tail;
head = tail = NULL;
for (int i = 0;i < 10;i++) {
listTailInsert(&head, &tail, a[i]);
}
listPrint(head);
DeleteCommon(&head);
listPrint(head);
return 0;
}
- 链表按奇数、偶数值拆分
#include<stdio.h>
typedef struct node
{
int val;
struct node *next;
}Node, *pNode;
void InsertTail(pNode *head, pNode *tail, int val) {
pNode newNode = (char*)calloc(1, sizeof(Node));
newNode->val = val;
if (NULL == *tail) {
*head = newNode;
*tail = newNode;
}
else {
(*tail)->next = newNode;
*tail = newNode;
}
}
void partition(pNode head, pNode *head1, pNode *tail1, pNode *head2, pNode *tail2) {
while (head) {
int t = head->val;
if (t % 2 == 1) {
InsertTail(head1, tail1, t);
}
else {
InsertTail(head2, tail2, t);
}
head = head->next;
}
}
void listPrint(pNode head) {
while (head) {
printf("%-3d", head->val);
head = head->next;
}
printf("\n");
}
int main() {
int a[9] = { 1,2,3,4,5,6,7,8,9 };
pNode head, tail, head1, tail1, head2, tail2;
head = tail = head1 = tail1 = head2 = tail2 = NULL;
for (int i = 0;i < 9;++i) {
InsertTail(&head, &tail, a[i]);
}
listPrint(head);
partition(head, &head1, &tail1, &head2, &tail2);
listPrint(head1);
listPrint(head2);
return 0;
}
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