前言
- 这个专栏将会用纯C实现常用的数据结构和简单的算法;
- 有C基础即可跟着学习,代码均可运行;
- 准备考研的也可跟着写,个人感觉,如果时间充裕,手写一遍比看书、刷题管用很多,这也是本人采用纯C语言实现的原因之一;
- 欢迎收藏 + 关注,本人将会持续更新。
双向链表
简介
🚡 双向链表,对比单链表来说,顾名思义,就是指向指向有两个指针,指向前后节点。
🌾 结合单链表,单链表有无头和有头之分,双向链表也是一样,这里是无头双链表,采用再封装写法,不是二级指针写法,再封装写法比较写起来比较容易,个人比较偏爱🤠🤠🤠🤠
双链表图示:
🉑 从ADT抽象中来说:
总结起来一句话:增删改查,🤠🤠🤠,假设双向链表是一个结合,这这个结合功能有:
- 增加元素
- 删除元素
- 拿取元素
- 查询元素
- 修改元素
- …………………………
双链表实现
这里采用再封装的方法,实现无头链表,有头和无头再单链表的那一节以及讲过了,这里采用无头实现
💠 节点封装
在双链表中,对于每一个节点来说,都有一个指向前、指向后节点的指针。
typedef int DataType;typedef struct Node {DataType data;struct Node* prev; // 前struct Node* next; // 后
}Node;
⚜️ 链表封装
这一步封装的作用是,可以更好的操作链表的一些操作,这个size的一个再封装写法的核心,无头与有头再实现的过程中,核心就在于第一个头节点的处理,无头如果没有任何节点,则插入的节点则作为第一个节点,但是这样会改变指针的指向,这也是因为需要传递二级指针的原因,而再封装写法则很好的解决了这个问题。
typedef struct List {Node* headNode;Node* tailNode;int size;
}List;
🍨 创建节点
采用calloc
创建节点,这样可以自动赋值为0
。
Node* create_node(DataType data)
{Node* node = (Node*)calloc(1, sizeof(Node));assert(node);node->data = data;return node;
}
🍤 创建链表
List* create_list()
{List* list = (List*)calloc(1, sizeof(List));assert(list);return list;
}
🤕 插入–头插
void push_front(List* list, DataType data)
{if (list == NULL) {return;}Node* node = create_node(data);if (list->size == 0) { // 这种写法的优势,不用传递二级指针list->tailNode = node;}else {node->next = list->headNode;list->headNode->prev = node;}list->headNode = node;list->size++;
}
🌮 插入–尾插入
- 情况判断:是否为空链表
- 为空:插入,作为头
- 不为空:找到尾节点,插入
void push_back(List* list, DataType data)
{if (list == NULL) {return;}Node* node = create_node(data);if (list->size == 0) {list->headNode = node;}else {list->tailNode->next = node;node->prev = list->tailNode;}list->tailNode = node;list->size++;
}
💠 插入–任意位置
- 规则: 在元素后位置插入
- 情况:3种
- 没有该元素
- 尾插
- 任意插
void insert(List* list, DataType posData, DataType data)
{if (list == NULL || list->size == 0) { // size != 0 保证有头return;}Node* t = list->headNode;while (t->next != NULL && t->data != posData) {t = t->next;}if (t->data != posData) { // 没有该元素return;}else if (t->next == NULL && t->data == posData) { // 尾插Node* node = create_node(data);node->prev = t;t->next = node;list->tailNode = node; // 尾巴移位}else{Node* node = create_node(data);node->next = t->next;t->next->prev = node;node->prev = t;t->next = node;}list->size++;
}
🎧 删除–头删
注意点:释放后指针指向问题:
- 如果说就一个元素,则删除后,在封装头需要指向NULL
- 如果不是,则下一个元素的prev指针需要赋值为NULL
void pop_front(List* list)
{if (list == NULL || list->size == 0) {return;}Node* node = list->headNode;list->headNode = node->next;free(node);(list->headNode) ? (list->headNode->prev = NULL) : (list->tailNode = NULL); // 判断是否只有一个节点的情况node = NULL;
}
🚖 删除–尾删除
注意点:释放后指针指向问题:
- 如果说就一个元素,则删除后,在封装头需要指向NULL
- 如果不是,则上一个元素的next指针需要赋值为NULL
void pop_back(List* list)
{if (list == NULL || list->size == 0) {return;}Node* node = list->tailNode;list->tailNode = list->tailNode->prev;free(node);(list->tailNode) ? (list->tailNode->next = NULL) : (list->headNode = NULL);list->size--;
}
🦅 删除–任意位置删除
四种情况:
- 没有找到
- 找到了
- 头删
- 尾删
- 任意位置删
void erase(List* list, DataType posData)
{if (list == NULL || list->size == 0) {return;}Node* cur = list->headNode;while (cur->next != NULL && cur->data != posData) {cur = cur->next;}// 没有找到if (cur->data != posData) {return;}else if (cur->next == NULL) { // 尾删除pop_back(list);}else if (cur->prev == NULL) { // 头删pop_front(list);}else {Node* t = cur;cur->prev->next = cur->next;cur->next->prev = cur->prev;free(t);t = NULL;list->size--;}
}
🌐 万金油函数
bool empty(List* list)
{if (list == NULL) {return true;}return list->size == 0;
}size_t size(List* list)
{if (list == NULL) {return 0;}return list->size;
}
📤 向前遍历
void travel_front(List* list)
{if (list == NULL) {return;}Node* cur = list->headNode;while (cur) {printf("%d ", cur->data);cur = cur->next;}printf("\n");
}
🎉 向后遍历
void travel_back(List* list)
{if (list == NULL) {return;}Node* cur = list->tailNode;while (cur) {printf("%d ", cur->data);cur = cur->prev;}printf("\n");
}
⏰ 总代码
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <assert.h>typedef int DataType;typedef struct Node {DataType data;struct Node* prev;struct Node* next;
}Node;typedef struct List {Node* headNode;Node* tailNode;int size;
}List;Node* create_node(DataType data)
{Node* node = (Node*)calloc(1, sizeof(Node));assert(node);node->data = data;return node;
}List* create_list()
{List* list = (List*)calloc(1, sizeof(List));assert(list);return list;
}void push_front(List* list, DataType data)
{if (list == NULL) {return;}Node* node = create_node(data);if (list->size == 0) {list->tailNode = node;}else {node->next = list->headNode;list->headNode->prev = node;}list->headNode = node;list->size++;
}void push_back(List* list, DataType data)
{if (list == NULL) {return;}Node* node = create_node(data);if (list->size == 0) {list->headNode = node;}else {list->tailNode->next = node;node->prev = list->tailNode;}list->tailNode = node;list->size++;
}void insert(List* list, DataType posData, DataType data)
{if (list == NULL || list->size == 0) { // size != 0 保证有头return;}Node* t = list->headNode;while (t->next != NULL && t->data != posData) {t = t->next;}if (t->data != posData) { // 没有该元素return;}else if (t->next == NULL && t->data == posData) { // 尾插Node* node = create_node(data);node->prev = t;t->next = node;list->tailNode = node; // 尾巴移位}else{Node* node = create_node(data);node->next = t->next;t->next->prev = node;node->prev = t;t->next = node;}list->size++;
}void pop_front(List* list)
{if (list == NULL || list->size == 0) {return;}Node* node = list->headNode;list->headNode = node->next;free(node);(list->headNode) ? (list->headNode->prev = NULL) : (list->tailNode = NULL); // 判断是否只有一个节点的情况node = NULL;
}void pop_back(List* list)
{if (list == NULL || list->size == 0) {return;}Node* node = list->tailNode;list->tailNode = list->tailNode->prev;free(node);(list->tailNode) ? (list->tailNode->next = NULL) : (list->headNode = NULL);list->size--;
}void erase(List* list, DataType posData)
{if (list == NULL || list->size == 0) {return;}Node* cur = list->headNode;while (cur->next != NULL && cur->data != posData) {cur = cur->next;}// 没有找到if (cur->data != posData) {return;}else if (cur->next == NULL) { // 尾删除pop_back(list);}else if (cur->prev == NULL) { // 头删pop_front(list);}else {Node* t = cur;cur->prev->next = cur->next;cur->next->prev = cur->prev;free(t);t = NULL;list->size--;}
}bool empty(List* list)
{if (list == NULL) {return true;}return list->size == 0;
}size_t size(List* list)
{if (list == NULL) {return 0;}return list->size;
}void travel_front(List* list)
{if (list == NULL) {return;}Node* cur = list->headNode;while (cur) {printf("%d ", cur->data);cur = cur->next;}printf("\n");
}void travel_back(List* list)
{if (list == NULL) {return;}Node* cur = list->tailNode;while (cur) {printf("%d ", cur->data);cur = cur->prev;}printf("\n");
}int main()
{List* list = create_list();for (int i = 1; i < 4; i++) {push_front(list, i);}for (int i = 1; i < 4; i++) {push_back(list, i * 10);}travel_front(list);travel_back(list);insert(list, 3, 33);insert(list, 30, 300);travel_front(list);travel_back(list);pop_front(list);travel_front(list);travel_back(list);pop_back(list);travel_front(list);travel_back(list);erase(list, 33);erase(list, 30);erase(list, 10);travel_front(list);travel_back(list);return 0;
}
循环链表
循环链表分为循环单链表,循环双链表,单链表和双链表又分为有头和无头链表,这里是有头循环双链表。
双向循环链表(Doubly Circular Linked List)是一种数据结构,其中每个节点都包含两个指针,一个指向前一个节点,一个指向后一个节点。与普通链表不同的是,双向循环链表的最后一个节点的下一个指针指向头节点,而头节点的前一个指针指向最后一个节点,形成一个循环。
🤕 节点封装
typedef int DataType;typedef struct Node {DataType data;struct Node* prev;struct Node* next;
}Node;
🍨 创建节点
Node* create_node(DataType data)
{Node* node = (Node*)calloc(1, sizeof(Node));assert(node);node->data = data;return node;
}
⚜️ 创建链表
这里需要构建一个循环节点(链表),如图:
Node* create_list()
{Node* list = (Node*)calloc(1, sizeof(Node));assert(list);list->next = list;list->prev = list;return list;
}
🎧 插入–头插
双向头删就很容易了,如图:
void push_front(Node* list, DataType data)
{assert(list);Node* node = create_node(data);node->next = list->next;list->next->prev = node; // 难点,不用担心 next,prev为空的时候node->prev = list;list->next = node;
}
🌮 插入–尾插
尾巴删除也很容易,因为头和尾巴互相找到,如图:
void push_back(Node* list, DataType data)
{assert(list);Node* node = create_node(data);node->prev = list->prev;list->prev->next = node;node->next = list;list->prev = node;
}
⛹️♀️ 插入–任意位置
任意位置也不难,找到要插入的位置,要注意的是找不到的情况。
// 找到要插入那个节点的位置节点
void insert(Node* list, DataType posData ,DataType data)
{assert(list);Node* cur = list->next;while (cur->next != list && cur->data != posData) {cur = cur->next;}if (cur->data != posData) { // 思考,为什么不能 cur->next != list ?????return;}else {Node* node = create_node(data);node->next = cur->next;cur->next->prev = node;node->prev = cur;cur->next = node;}}
👟 删除–头删
注意: 一个节点的头节点指向不同。
两种情况:
- 如果一个元素,这个时候删除,头节点指向修改和两个元素以上删除不同,这个时候头节点需要指向自己
- 两个元素及上
void pop_front(Node* list)
{assert(list);Node* cur = list->next;if (cur == list) { // 无节点return;}else if(cur->next == list) { // 一个节点list->prev = list;list->next = list;}else {list->next = cur->next; // 两个节点以上cur->next->prev = list;}free(cur);cur = NULL;
}
📑 删除–尾删
这个也是简单的,因为可以通过头节点直接找到尾节点,这个时候就只需要一种情况即可,因为创建双链表有一个很好的特性,
void pop_back(Node* list)
{assert(list);Node* cur = list->prev; // 因为可以获取尾节点if (cur == list) {return;}else {cur->prev->next = list; // 哪怕是一个节点,也和普通情况也是一样list->prev = cur->prev; // 这个也是一样,free(cur);cur = NULL;}}
🚴♀ 删除–任意位置
很简单,因为这个也不用记录前驱节点,也不用找尾节点了,只需要考虑两种情况:
- 没有找到
- 找到了
void erase(Node* list, DataType posData)
{assert(list);Node* cur = list->next;while (cur->next != list && cur->data != posData) {cur = cur->next;}if (cur->data != posData) {return;}else {cur->prev->next = cur->next;cur->next->prev = cur->prev;free(cur);cur = NULL;}
}
🔱 遍历
void travel_front(Node* list)
{assert(list);Node* cur = list->next;while (cur != list) {printf("%d ", cur->data);cur = cur->next;}printf("\n");
}void travel_back(Node* list)
{assert(list);Node* cur = list->prev;while (cur != list) {printf("%d ", cur->data);cur = cur->prev;}printf("\n");
}
⚗️ 总代码
#include <stdlib.h>
#include <assert.h>
#include <stdbool.h>
#include <stdio.h>// 有头链表实现,简单点typedef int DataType;typedef struct Node {DataType data;struct Node* prev;struct Node* next;
}Node;Node* create_node(DataType data)
{Node* node = (Node*)calloc(1, sizeof(Node));assert(node);node->data = data;return node;
}Node* create_list()
{Node* list = (Node*)calloc(1, sizeof(Node));assert(list);list->next = list;list->prev = list;return list;
}void push_front(Node* list, DataType data)
{assert(list);Node* node = create_node(data);node->next = list->next;list->next->prev = node; // 难点,不用担心 next,prev为空的时候node->prev = list;list->next = node;
}void push_back(Node* list, DataType data)
{assert(list);Node* node = create_node(data);node->prev = list->prev;list->prev->next = node;node->next = list;list->prev = node;
}
// 找到要插入那个节点的位置节点
void insert(Node* list, DataType posData ,DataType data)
{assert(list);Node* cur = list->next;while (cur->next != list && cur->data != posData) {cur = cur->next;}if (cur->data != posData) { // 思考,为什么不能 cur->next != list ?????return;}else {Node* node = create_node(data);node->next = cur->next;cur->next->prev = node;node->prev = cur;cur->next = node;}}void pop_front(Node* list)
{assert(list);Node* cur = list->next;if (cur == list) {return;}else if(cur->next == list) {list->prev = list;list->next = list;}else {list->next = cur->next;cur->next->prev = list;}free(cur);cur = NULL;
}void pop_back(Node* list)
{assert(list);Node* cur = list->prev;if (cur == list) {return;}else {cur->prev->next = list;list->prev = cur->prev;free(cur);cur = NULL;}}void erase(Node* list, DataType posData)
{assert(list);Node* cur = list->next;while (cur->next != list && cur->data != posData) {cur = cur->next;}if (cur->data != posData) {return;}else {cur->prev->next = cur->next;cur->next->prev = cur->prev;free(cur);cur = NULL;}
}void travel_front(Node* list)
{assert(list);Node* cur = list->next;while (cur != list) {printf("%d ", cur->data);cur = cur->next;}printf("\n");
}void travel_back(Node* list)
{assert(list);Node* cur = list->prev;while (cur != list) {printf("%d ", cur->data);cur = cur->prev;}printf("\n");
}int main()
{Node* list = create_list();push_front(list, 1);push_front(list, 2);push_front(list, 3);travel_front(list);travel_back(list);push_back(list, 11);push_back(list, 22);push_back(list, 33);travel_front(list);travel_back(list);insert(list, 2, 20);insert(list, 3, 30);insert(list, 33, 330);travel_front(list);travel_back(list);pop_front(list);travel_front(list);travel_back(list);pop_back(list);travel_front(list);travel_back(list);erase(list, 33);travel_front(list);travel_back(list);return 0;
}
循环链表约瑟夫环问题
讲一个比较有意思的故事:约瑟夫是犹太军队的一个将军,在反抗罗马的起义中,他所率领的军队被击溃,只剩下残余的部队40余人,他们都是宁死不屈的人,所以不愿投降做叛徒。一群人表决说要死,所以用一种策略来先后kill所有人。
于是约瑟夫建议:每次由其他两人一起kill一个人,而被kill的人的先后顺序是由抽签决定的,约瑟夫有预谋地抽到了最后一签,在kill了除了他和剩余那个人之外的最后一人,他劝服了另外一个没死的人投降了罗马。
我们这个规则是这么定的:
- 在一间房间总共有n个人(下标0~n-1),只能有最后一个人活命。
- 按照如下规则去排除人:
- 所有人围成一圈
- 顺时针报数,每次报到q的人将被排除掉
- 被排除掉的人将从房间内被移走
- 然后从被kill掉的下一个人重新报数,继续报q,再清除,直到剩余一人
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>/*
* 用上一个链表,也可以。这里采用无头循环双链表实现
* 无头采用再次封装的写法
*/typedef struct Node {int data;struct Node* prev;struct Node* next;
}Node;typedef struct List {Node* headNode;
}List;// 每一个节点创建都是循环
Node* create_node(int data)
{Node* node = (Node*)calloc(1, sizeof(Node));assert(node);node->data = data;node->prev = node;node->next = node;return node;
}void push_back(List* list, int data)
{assert(list);Node* node = create_node(data);if (list->headNode == NULL) {list->headNode = node;}else {Node* cur = list->headNode->prev;node->next = list->headNode;list->headNode->prev = node;cur->next = node;node->prev = cur;}
}void erase(List* list, Node* node)
{assert(list);assert(node);// 一个节点if (node->next == node) {free(node);node = NULL;list->headNode = NULL;}else {node->prev->next = node->next;node->next->prev = node->prev;if (list->headNode == node) { // 防止删除头list->headNode = node->next;}free(node);node = NULL;}
}void travel(List* list)
{Node* cur = list->headNode;while (cur->next != list->headNode) {printf("%d ", cur->data);cur = cur->next;}printf("%d ", cur->data);printf("\n");
}// 只做演示,不考虑内存问题
void game_run(int n, int m)
{if (n < 0 || m < 0) {return;}List list = { NULL };for (int i = 1; i <= n; i++) {push_back(&list, i);}travel(&list);Node* cur = list.headNode;while (n > 1) {// 报数for (int i = 1; i < m; i++) {cur = cur->next;}Node* next = cur->next;erase(&list, cur);// 重置报数cur = next;n--;}printf("天选之子: %d\n", cur->data);
}int main()
{game_run(10, 3);return 0;
}