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数据结构研究如何使用存储区解决问题
算法研究解决常见问题的方法

数字之间的关系可以从两个完全不同的角度
进行描述
逻辑关系（逻辑结构）描述数字之间和计算机
无关的关系
物理关系（物理结构）描述存放数字的存储区
之间的关系

逻辑结构分为如下几种
1.集合结构：所有数字可以看作一个整体
2.线性结构：可以用一条有顺序的线把所有
数字连起来
3.树状结构：所有数据都是从一个数据开始
向一个方向扩展出来的，任何数据
可以扩展出多个其他数据
4.网状结构：任何两个数字之间可以有直接的
联系，所有数字之间的联系没有统一
方向

物理结构有以下两种
1.顺序结构：所有存储区在内存里连续排列
数组和动态分配内存都是顺序结构的例子
顺序结构中每个存储区有一个编号，
可以根据编号直接找到对应的存储区
根据编号找到存储区的方法叫随机访问，
顺序结构支持随机访问能力
顺序结构中存储区个数很难调整，这
有可能造成内存的浪费
顺序结构不适合进行插入或删除操作

/*顺序结构演示
*/
#include <stdio.h>
void remove_num(int *p_num, int size, int num) {int num1 = 0;for (num1 = 0;num1 <= size - 1;num1++) {if (num < *(p_num + num1)) {*(p_num + num1 - 1) = *(p_num + num1);}else if (num1 && *(p_num + num1) < *(p_num + num1 - 1)) {*(p_num + num1 - 1) = *(p_num + num1);break;}}
}
void insert(int *p_num, int size, int num) {int tmp = num, num1 = 0, tmp1 = 0;for (num1 = 0;num1 <= size - 1;num1++) {if (tmp < *(p_num + num1)) {tmp1 = tmp;tmp = *(p_num + num1);*(p_num + num1) = tmp1;}else if (num1 && *(p_num + num1) < *(p_num + num1 - 1)) {tmp1 = tmp;tmp = *(p_num + num1);*(p_num + num1) = tmp1;break;}}
}
int main() {int arr[20] = {2, 6, 12, 14, 17, 21, 23,31, 35, 37};int num = 0;insert(arr, 20, 27);for (num = 0;num <= 19;num++) {printf("%d ", arr[num]);}printf("\n");remove_num(arr, 20, 14);for (num = 0;num <= 19;num++) {printf("%d ", arr[num]);}printf("\n");return 0;
}

 

2.链式物理结构：由多个无关的存储区构成，
任何两个存储区之间可以用指针连接

链式物理结构中每个存储区叫做一个
结点
单向线性链式物理结构中任何两个结点
之间都有前后关系（每个结点里只
需要包含一个指针）
单向线性链式物理结构中最后一个结点
里的指针必须是空指针
链式物理结构不直接支持随机访问能力
可以在所有结点前增加一个无效头结点，
在所有结点后增加一个无效尾结点，这样
可以简化程序的编写　　

/*链式物理结构演示
*/
#include <stdio.h>
typedef struct node {int num;struct node *p_next;
} node;
int main() {node node1 = {1}, node2 = {5}, node3 = {12}, head = {0}, tail = {0}, node4 = {7};int cnt = 0;node *p_node = NULL;node1.p_next = &node2;node2.p_next = &node3;head.p_next = &node1;node3.p_next = &tail;for (p_node = &head;p_node != &tail;p_node = p_node->p_next) {node *p_first = p_node;node *p_mid = p_first->p_next;node *p_last = p_mid->p_next;if (p_mid != &tail) {printf("%d ", p_mid->num);}}printf("\n");for (p_node = &head;p_node != &tail;p_node= p_node->p_next) {node *p_first = p_node;node *p_mid = p_first->p_next;node *p_last = p_mid->p_next;if (p_mid != &tail && cnt == 2) {printf("数字是%d\n", p_mid->num);}cnt++;}for (p_node = &head;p_node != &tail;p_node = p_node->p_next) {node *p_first = p_node;node *p_mid = p_first->p_next;node *p_last = p_mid->p_next;if (p_mid == &tail || p_mid->num > node4.num) {p_first->p_next = &node4;node4.p_next = p_mid;break;}}for (p_node = &head;p_node != &tail;p_node = p_node->p_next) {node *p_first = p_node;node *p_mid = p_first->p_next;node *p_last = p_mid->p_next;if (p_mid != &tail) {printf("%d ", p_mid->num);}}printf("\n");for (p_node = &head;p_node != &tail;p_node = p_node->p_next) {node *p_first = p_node;node *p_mid = p_first->p_next;node *p_last = p_mid->p_next;if (p_mid != &tail && p_mid->num == 5) {p_first->p_next = p_last;break;}}for (p_node = &head;p_node != &tail;p_node = p_node->p_next) {node *p_first = p_node;node *p_mid = p_first->p_next;node *p_last = p_mid->p_next;if (p_mid != &tail) {printf("%d ", p_mid->num);}}printf("\n");return 0;
}

 

链式物理结构中每个有效结点都应该是动态
分配的
链式物理结构中能容纳的数字数量可以灵活
变化

数据结构由一组存储区和一组相关的函数构成
这些函数提供了对存储区的使用方法
程序中的其他语句只能通过这组函数使用这些
存储区

/*动态分配链式物理结构演示
*/
#include <stdio.h>
#include <stdlib.h>
typedef struct node {int num;struct node *p_next;
} node;
int main() {int num = 0;node head = {0}, tail = {0}, *p_tmp = NULL, *p_node = NULL;head.p_next = &tail;while (1) {printf("请输入一个数字：");scanf("%d", &num);if (num < 0) {break;}p_tmp = (node *)malloc(sizeof(node));if (!p_tmp) {continue;}p_tmp->num = num;p_tmp->p_next = NULL;for (p_node = &head;p_node != &tail;p_node = p_node->p_next) {node *p_first = p_node;node *p_mid = p_first->p_next;node *p_last = p_mid->p_next;if (p_mid == &tail) {p_first->p_next = p_tmp;p_tmp->p_next = p_mid;break;}}}printf("请输入要删除的数字：");scanf("%d", &num);for (p_node = &head;p_node != &tail;p_node = p_node->p_next) {node *p_first = p_node;node *p_mid = p_first->p_next;node *p_last = p_mid->p_next;if (p_mid != &tail && p_mid->num == num) {p_first->p_next = p_last;free(p_mid);p_mid = NULL;break;}}for (p_node = &head;p_node != &tail;p_node = p_node->p_next) {node *p_first = p_node;node *p_mid = p_first->p_next;node *p_last = p_mid->p_next;if (p_mid != &tail) {printf("%d ", p_mid->num);}}printf("\n");while (head.p_next != &tail) {node *p_first = &head;node *p_mid = p_first->p_next;node *p_last = p_mid->p_next;p_first->p_next = p_last;free(p_mid);p_mid = NULL;}return 0;
}