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逆波兰表达式:
计算给定的逆波兰表达式的值,有效操作只有加减乘除,每个操作数都为整数。
如:
"2","1","+","3","*" : 9;---------(2+1)*3
"4","13","5","/","+" : 6;---------4+(13/5)
程序实现:
1 #include <iostream> 2 #include <stack> 3 #include <algorithm> 4 #include <cstring> 5 using namespace std; 6 //判断是否为运算符 7 bool IsOperator(const char* token){ 8 return ((token[0]==‘+‘)||(token[0]==‘-‘) 9 ||(token[0]==‘*‘)||(token[0]==‘/‘)); 10 } 11 //逆波兰表达式,利用压栈出栈的方式: 12 /* 13 若当前字符为操作数,则压栈 14 若当前字符为操作符,则弹出栈中的两个操作数,进行计算后再次压入栈中 15 */ 16 int ReversePolishNotation(const char* str[],int size){ 17 stack<int> s; 18 int a,b; 19 const char* token; 20 for(int i=0;i<size;i++){ 21 token = str[i]; 22 if(!IsOperator(token)) 23 s.push(atoi(token)); 24 else{ 25 b = s.top(); 26 s.pop(); 27 a = s.top(); 28 s.pop(); 29 if(token[0] == ‘+‘) 30 s.push(a+b); 31 else if(token[0] == ‘-‘) 32 s.push(a-b); 33 else if(token[0] == ‘*‘) 34 s.push(a*b); 35 else if(token[0] == ‘/‘) 36 s.push(a/b); 37 } 38 } 39 return s.top(); 40 } 41 int main() 42 { 43 const char* str[] = {"2","1","+","3","*"}; 44 const char* str1[] = {"4","13","5","/","+"}; 45 int value = ReversePolishNotation(str,sizeof(str)/sizeof(const char*)); 46 cout<<value<<endl; 47 cout<<ReversePolishNotation(str1,sizeof(str1)/sizeof(const char*))<<endl; 48 return 0; 49 }
运行结果:
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C++博客园:godfrey_88
http://www.cnblogs.com/gaobaoru-articles/
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原文地址:http://www.cnblogs.com/gaobaoru-articles/p/5448977.html