数据结构——栈的应用

一.括号匹配的检验

遇到左括号就进栈，遇到右括号就跟当前栈顶匹配，能匹配就将其出栈，否则不匹配

 

#include <stdio.h>
#include <malloc.h>

#define STACK_MAX_SIZE 10

//创建栈结构体
typedef struct charStack {
	int top;

	int data[STACK_MAX_SIZE];
} *stackPtr;

//输出栈
void outputStack(stackPtr paraStack) {
	for (int i = 0; i <= paraStack->top; i++) {
		printf("%d ", paraStack->data[i]);
	}
	printf("\r\n");
}

//初始化栈
stackPtr initStack() {
	stackPtr tempStack = (stackPtr)malloc(sizeof(stackPtr));
	tempStack->top = -1;

	return tempStack;
}

//入栈
void push(stackPtr stack, int paraValue) {
	//检查空间
	if (stack->top >= STACK_MAX_SIZE - 1) {
		printf("Cannot push element: stack full.\r\n");
		return;
	}

	//update top
	stack->top++;

	//添加元素
	stack->data[stack->top] = paraValue;
}

//出栈
int pop(stackPtr stack) {
	// 空间检查
	if (stack->top < 0) {
		printf("Cannot pop element: stack empty.\r\n");
		return '\0';
	}

	//update top
	stack->top--;

	//返回栈顶
	return stack->data[stack->top + 1];
}

bool bracketMatch(char *paraString, char paraLength) {
	//初始化创建栈底
	stackPtr tempStack = initStack();
	push(tempStack, '#');
	char tempChar, tempPopedChar;

	//start
	for (int i = 0; i < paraLength; i++) {
		tempChar = paraString[i];

		switch (tempChar) {
			case '(':
			case '[':
			case '{':
				push(tempStack, tempChar);
				break;
			case ')':
				tempPopedChar = pop(tempStack);
				if (tempPopedChar != '(') {
					return false;
				}
				break;
			case ']':
				tempPopedChar = pop(tempStack);
				if (tempPopedChar != '[') {
					return false;
				}
				break;
			case '}':
				tempPopedChar = pop(tempStack);
				if (tempPopedChar != '{') {
					return false;
				}
				break;
			default:
				break;
		}
	}

	tempPopedChar = pop(tempStack);
	if (tempPopedChar != '#') {
		return false;
	}
	return true;
}


void pushPopTest() {
	printf("---- pushPopTest begins. ----\r\n");

	// Initialize.
	stackPtr tempStack = initStack();
	printf("After initialization, the stack is: ");
	outputStack(tempStack);

	// Pop.
	for (int i = 1; i < 10; i++) {
		printf("Pushing %d.\r\n", i);
		push(tempStack, i);
		outputStack(tempStack);
	}//Of for i

	// Pop.
	int s;
	for (int j = 0; j < 3; j++) {
		s = pop(tempStack);
		printf("Pop %d.\r\n", s);
		outputStack(tempStack);
	}//Of for i

	printf("---- pushPopTest ends. ----\r\n");
}// Of pushPopTest

void bracketMatchingTest() {
	char *tempExpression = "[2 + (1 - 3)] * 4";
	bool tempMatch = bracketMatch(tempExpression, 17);
	printf("Is the expression '%s' bracket matching? %d \r\n", tempExpression, tempMatch);


	tempExpression = "( () )";
	tempMatch = bracketMatch(tempExpression, 6);
	printf("Is the expression '%s' bracket matching? %d \r\n", tempExpression, tempMatch);

	tempExpression = "（()())";
	tempMatch = bracketMatch(tempExpression, 6);
	printf("Is the expression '%s' bracket matching? %d \r\n", tempExpression, tempMatch);

	tempExpression = "({}[])";
	tempMatch = bracketMatch(tempExpression, 6);
	printf("Is the expression '%s' bracket matching? %d \r\n", tempExpression, tempMatch);


	tempExpression = "[(])";
	tempMatch = bracketMatch(tempExpression, 2);
	printf("Is the expression '%s' bracket matching? %d \r\n", tempExpression, tempMatch);
}// Of bracketMatchingTest


/**
 The entrance.
 */
int main() {
	//pushPopTest();
	bracketMatchingTest();
}

二.表达式求值（此部分多数是抄的）

第一种方法代码如下

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h> 

#define Maxsize 100

typedef int dataType; 
typedef struct Stack
{
	dataType *top;
	dataType *base;
	int stacksize;
}sqstack;

void create(sqstack *s)
{
    s->base=(dataType *)malloc(Maxsize*sizeof(dataType)); 
     if(!s->base)
     {
    	printf("Space allocation failed!\n");
    	return;
	 }
    s->top=s->base;
    s->stacksize=Maxsize;
    return;
}

int push_in(sqstack *s,dataType value)
{
	if(s->top-s->base==s->stacksize)
	{
		printf("Stack is full!\n");
		return;
	}
	*s->top=value; //"*s->top++=value;"
	s->top++;
	return;
}

int pop_out(sqstack *s,dataType *elem)
{
	if(s->base==s->top)
	{
		printf("Stack is empty!");
		return;
	}
	s->top--;
	*elem=*s->top;  //"*elem=*--s->top;"
	return;
}

dataType GetTop(sqstack *s) 
{
	if(s->base==s->top)
	{
	    printf("Stack is empty!\n");
	    printf("Unable to fetch top stack element!\n");
	    return;
	} 
	return *(s->top-1);
}

char Precede(char theta1,char theta2)
{
	int i,j;
	char pre[7][7]={// +   -   *   /   (   )   = 
		             {'>','>','<','<','<','>','>'},  
                     {'>','>','<','<','<','>','>'},  
                     {'>','>','>','>','<','>','>'},  
                     {'>','>','>','>','<','>','>'},  
                     {'<','<','<','<','<','=','0'},  
                     {'>','>','>','>','0','>','>'},  
                     {'<','<','<','<','<','0','='}}; 
                
    switch(theta1){
    	case '+': i=0; break;  
        case '-': i=1; break;  
        case '*': i=2; break;  
        case '/': i=3; break;  
        case '(': i=4; break;  
        case ')': i=5; break;
        case '=': i=6; break;
	}
	
    switch(theta2){
    	case '+': j=0; break;  
        case '-': j=1; break;  
        case '*': j=2; break;  
        case '/': j=3; break;  
        case '(': j=4; break;  
        case ')': j=5; break;
        case '=': j=6; break;
    }
    return(pre[i][j]);
}

int Operate(int a,char theta,int b)
{
	int result;
	
	switch(theta){
		case'+':return a+b;
		case'-':return a-b;
		case'*':return a*b;
		case'/':             //判断除数是否为0，若除数为零返回错误提示                       
		        if(b!=0)
		        return a/b;
		        else
		        {
		        	printf("Divisor can not Be zero!\n");
		        	exit(0);
				}
	}
    
}

int In(char c) //判断是否为运算符，是运算符返回1，若不是返回0
{
	switch(c){
		case '+':   
        case '-':  
        case '*':   
        case '/':   
        case '(':   
        case ')': 
        case '=': 
                 return 1;
		 default:
		         return 0;		   
	}
}

int evaluateExpression(char exp[]) 
{
	sqstack OPND,OPTR; //运算数栈、运算符栈
	dataType a,b,theta,x,X1,X2;  /* a，b，theta用于Operate函数
	                                X用于存放多余的出栈字符
	                                X1,X2用于归并 */
	char ch; //读取字符的变量
	int i=0; //指向存放表达式数组的下标指针（其实不是真正的指针，而是数组下标）
	
	create(&OPND); //建立并初始化运算数栈OPND
	create(&OPTR); //建立并初始化运算符栈OPTR
	push_in(&OPTR,'='); //先将“=”号压入OPTR栈（表达式也必须以“=”结束）
	
	ch=exp[i++]; //ch读取字符数组第一个元素，并将指针i后移一位

	
	while(ch!='='||GetTop(&OPTR)!='=') //表达式没有扫描完毕或OPTR的栈顶不为“=”
	{
		if(In(ch)) //判断ch是否为操作符
		{
			switch(Precede(GetTop(&OPTR),ch))//比较ch的与OPTR栈顶元素的优先级
			{
				case'<':
					    push_in(&OPTR,ch);
					    ch=exp[i++]; //读取下一位字符并将指针向后偏移一位
					    break;
				
				case'>':
					    pop_out(&OPTR,&theta); 
					    pop_out(&OPND,&b);
					    pop_out(&OPND,&a);
					    push_in(&OPND,Operate(a,theta,b));
					    break;
				
				case'=':
					    pop_out(&OPTR,&x);
					    ch=exp[i++]; //读取下一位字符并将指针向后偏移一位
					    break;
			}
		
		}
		
		else if(isdigit(ch)) //判断是否为数字
		{
			X1=ch-'0'; //将字符形式转化为数字
			push_in(&OPND,X1);
			X2=X1;
			ch=exp[i++]; //读取下一位字符并将指针向后偏移一位
			
			while(isdigit(ch)) //判断下一位是否还是数字
			{
				X1=ch-'0';
				X2=10*X2+X1; //归并至X2
				pop_out(&OPND,&x);
				push_in(&OPND,X2);
				ch=exp[i++]; //读取下一位字符并将指针向后偏移一位
				
			} 
		} 
		
		else if(ch==' ') //判断是否为空格
		{
			while(ch==' ') //跳过若干个空格
			{
				ch=exp[i++]; //忽略空格，直接取下一位
			}
			
		}
		
		else //若不是上述三种情况之一，则为非法字符
		{
			printf("Input has illegal characters!\n");
			printf("Please enter again.\n");
			exit(0); //返回错误提示
		}
	}
	return(GetTop(&OPND)); //最后返回操作数栈顶为运算结果
}

int main()
{

  char exp[100]; //定义一个字符数组用于存储表达式
  int result;
  printf("Please Enter Expression:");
  gets(exp); //换成scanf（“%s”，exp）；也可以
  result=evaluateExpression(exp);
  printf("\n");
  printf("%s%d\n",exp,result);
  
	return 0;
}

还有一种方法是这样

 但是建立两个不同类型的栈，需要在代码中将栈的基本操作函数（建立，初始化，入栈，出栈，取栈顶元素）写两遍，然后在存取操作数和运算符的时候分别调用对应的操作，过程过于麻烦，代码也过于繁琐。