Day04_数据结构(难点)

01.思维导图

02.地址传递和值传递

#include <stdio.h>
void swap(int **p1,int **p2)
{int *temp;temp = *p1;*p1 = *p2;*p2 = temp;
}
void swap2(int *p1,int *p2)
{int temp;temp = *p1;*p1 = *p2;*p2 = temp;
}int main(int argc, const char *argv[])
{int a = 90,b=7;int *p1 = &a,*p2=&b;swap(&p1,&p2);swap2(p1,p2);printf("a=%d b=%d",a,b);                return 0;}

输出结果：a=7,b=90

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int swap(int **p1,int **p2)
{// 将 p1 指向的变量的值加 1(**p1)++;// 将 p1 指向的变量的值赋给 p2 指向的变量**p2 = **p1;// 将 p2 指向的变量的值加上 p1 指向的变量的值**p2 = **p2 + **p1;return **p2;                                         
}   
int main(int argc, const char *argv[])
{int a=90,b=7;int *p1=&a,*p2=&b;printf("%d\n",swap(&p1,&p2));printf("a=%d\tb=%d\n",a,b);return 0;
}   

03.栈和递归的联系

利用今天实现的栈，完成输出一个数的二进制。可以利用条件判断，完成指定进制的输出

main.c

#include "binary.h"
int main()
{int number=255;int bases[]={2,8,16};int num_bases=sizeof(bases)/sizeof(bases[0]);for(int i=0;i<num_bases;i++){printf("%d 的 %d进制是:",number,bases[i]);converToBinary(number,bases[i]);}return 0;
}

binary.c

#include "binary.h"
//1.创建栈
stack_p createstack(int capacity)
{stack_p S=(stack_p)malloc(sizeof(stack));S->items=(int*)malloc(capacity*sizeof(int));S->top=-1;S->capacity=capacity;return S;
}
//2.判空
int empty_stack(stack_p S)
{if(S==NULL){printf("入参为空.\n");return -1;}return S->top==-1?1:0;
}
//3.入栈操作
void push(stack_p S,int item)
{if(S==NULL){printf("入参为空.\n");return;}S->items[++S->top]=item;
}
//4.出栈操作
int pop(stack_p S)
{if(S==NULL){printf("入参为空.\n");return -1;}if(empty_stack(S)){printf("空栈无法输出.\n");return -2;}return S->items[S->top--];
}
void converToBinary(int num,int base)
{if(num==0){printf("0\n");return;}//假设最大32位整数stack_p S=createstack(32);char digits[]="0123456789ABCDEF";while(num>0){int remainder=num%base;push(S,remainder);num/=base;}while(!empty_stack(S)){int digit=pop(S);printf("%c",digits[digit]);}printf("\n");free(S->items);free(S);
}

binary.h

#ifndef __BINARY_H__
#define __BINARY_H__
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef struct stack
{int *items;int top;int capacity;}stack,*stack_p;
stack_p createstack(int capacity);
int empty_stack(stack_p S);
void push(stack_p S,int item);
int pop(stack_p S);
void converToBinary(int num,int base);#endif

2. 栈模拟递归计算阶乘

#include <stdio.h>
#include <stdlib.h>// 定义栈结构
typedef struct Stack {int *items;int top;int capacity;
} Stack;// 创建栈
Stack* createStack(int capacity) {Stack* stack = (Stack*)malloc(sizeof(Stack));stack->items = (int*)malloc(capacity * sizeof(int));stack->top = -1;stack->capacity = capacity;return stack;
}// 判断栈是否为空
int isEmpty(Stack* stack) {return stack->top == -1;
}// 入栈操作
void push(Stack* stack, int item) {stack->items[++stack->top] = item;
}// 出栈操作
int pop(Stack* stack) {if (isEmpty(stack)) {return -1;}return stack->items[stack->top--];
}// 释放栈内存
void freeStack(Stack* stack) {free(stack->items);free(stack);
}// 使用栈模拟递归计算阶乘
int factorial_stack(int n) {Stack* stack = createStack(n);int result = 1;// 将参数依次压入栈while (n > 0) {push(stack, n);n--;}// 从栈中弹出元素并计算阶乘while (!isEmpty(stack)) {result *= pop(stack);}freeStack(stack);return result;
}// 递归计算阶乘
int factorial_recursive(int n) {if (n == 0 || n == 1) {return 1;} else {return n * factorial_recursive(n - 1);}
}int main() {int num = 5;printf("使用栈计算 %d 的阶乘: %d\n", num, factorial_stack(num));printf("使用递归计算 %d 的阶乘: %d\n", num, factorial_recursive(num));return 0;
}

3.递归调用过程中系统栈的使用：斐波那契数列

#include <stdio.h>
#include <stdlib.h>// 定义栈结构
typedef struct Stack {int *items;int top;int capacity;
} Stack;// 创建栈
Stack* createStack(int capacity) {Stack* stack = (Stack*)malloc(sizeof(Stack));stack->items = (int*)malloc(capacity * sizeof(int));stack->top = -1;stack->capacity = capacity;return stack;
}// 判断栈是否为空
int isEmpty(Stack* stack) {return stack->top == -1;
}// 入栈操作
void push(Stack* stack, int item) {stack->items[++stack->top] = item;
}// 出栈操作
int pop(Stack* stack) {if (isEmpty(stack)) {return -1;}return stack->items[stack->top--];
}// 释放栈内存
void freeStack(Stack* stack) {free(stack->items);free(stack);
}// 使用栈计算斐波那契数列的第 n 项
int fibonacciStack(int n) {if (n == 0) return 0;if (n == 1) return 1;Stack* stack = createStack(n);// 先将 n 压入栈push(stack, n);int result = 0;while (!isEmpty(stack)) {int current = pop(stack);if (current == 0) {result += 0;} else if (current == 1) {result += 1;} else {// 模拟递归调用，先压入较小的参数push(stack, current - 1);push(stack, current - 2);}}freeStack(stack);return result;
}int main() {int n = 6;printf("斐波那契数列的第 %d 项是: %d\n", n, fibonacciStack(n));return 0;
}

课堂重点