C语言强化——链表（1）

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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C语言强化——链表（1）

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