數據結構08——逆波蘭式

Time Limit: 3000ms, Memory Limit: 10000KB , Accepted: 277, Total Submissions: 763

Description

假設表達式由單字母變量和雙目四則運算算符構成。試編寫程序，將一個通常書寫形式且書寫正確的表達式轉換爲逆波蘭式。

Input

輸入由單字母變量和雙目四則運算算符構成的表達式。

Output

輸出其逆波蘭式。

• Sample Input 

  (a+b)*c

• Sample Output

  ab+c*

• #include<stdio.h>
  #include<stdlib.h>
  
  typedef struct stack{
  	struct stack *pop;
  	struct stack *bottom;
  	struct stack *next;
  	char sign;
  }stack; 
  
  void init(stack *list){
      list->bottom = (stack*) malloc (sizeof(stack));
  	list->bottom->next = NULL;
  	list->pop = list->bottom;
  }
  
  void push(stack *list, char s){
      stack *p;
  	p = (stack*)malloc(sizeof(stack));
  	p->sign = s;
  	p->next = list->pop;
  	list->pop = p;
  }
  
  bool judge(char a, char b){
      if((a == '*' && (b == '+' || b == '-')) || (a == '/' && (b == '+' || b == '-')))
  		return true;
  	else
  		return false;
  }
  
  void pop(stack *list){
      stack *p;
  	p = list->pop;
  	list->pop = list->pop->next;
  	free(p);
  }
  
  int main(){
  	char s,t;
  	stack *list;
  	list = (stack*) malloc (sizeof(stack));
  	init(list);
  	while(scanf("%c",&s) && s!='\n'){
  		if(s >= 'a' && s <= 'z'){
  		    printf("%c",s);
  		}
  		else if(s == '('){
  		    push(list,s);
  		}
  		else if(s == ')'){
  			while(list->pop->sign != '('){
  				printf("%c",list->pop->sign);
  			    pop(list);
  			}
  			pop(list);
  		}
  		else{
  			while(1){
  				t=list->pop->sign;
  			    if(judge(s, list->pop->sign) || list->pop == list->bottom || list->pop->sign == '('){
  		            push(list, s);
  					break;
  		        }
  				else if(t == '+' || t == '-' || t == '*' || t == '/'){
  				    printf("%c",list->pop->sign);
  			        pop(list);
  				}
  				else{
  					break;
  				}
  			}   
  		}
  	}
      while(list->pop != list->bottom){
      	printf("%c",list->pop->sign);
  		pop(list);
      }
      printf("\n");
  	return 0;
  }

  
  
  

複習：

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

typedef struct node{
	char data;
	struct node *next;
}node, *pnode;

typedef struct stack{
	struct node *bottom;
	struct node *top;
}stack, *pstack;

void init(stack *list){
	node *p = (pnode)malloc(sizeof(node));
	p->data = '#';
	p->next = NULL;
	list->bottom = p;
	list->top = p;
}

void push(stack *list, char x){
    node *p = (pnode)malloc(sizeof(node));
	p->data = x;
	p->next = NULL;
	list->top->next = p;
	list->top = p;
}

bool isempty(stack *list){
	if(list->top == list->bottom)
		return true;
	else
		return false;
}

void pop(stack *list){
	pnode p, q;
	p = list->top;
    q = list->bottom;
	while(q->next != p)
		q = q->next;
	list->top = q;
	free(p);
}

bool judgelow(pstack list, char x){
	char y;
	y = list->top->data;
    if((x == '*' ||x == '/') && (y == '+' || y =='-'))
		return false;
	else
		return true;
}

int main(){
    char s;
	pstack list = (pstack)malloc(sizeof(stack));
	pnode p;
	init(list);
	while(scanf("%c", &s)){
	    if(s == '\n')
			break;
		else if(s >= 'a' && s <= 'z')
			printf("%c", s);
		else if(s == '(')
			push(list, s);
		else if(s == ')'){
			while(list->top->data != '('){
				printf("%c", list->top->data);
		        pop(list);
			}
			pop(list);
    	}
		else{
			while(1){
				if(!judgelow(list, s) || isempty(list) || list->top->data == '('){
                    push(list, s);
				    break;
				}
				else{
			        printf("%c", list->top->data);
				    pop(list);
				}
			}
			
		}
	}
	while(!isempty(list)){
	    printf("%c", list->top->data);
		pop(list);
	}
	return 0;
}