c++实现将简单的中缀表达式转化为…

来源：互联网发布：jenkins php持续集成编辑：程序博客网时间：2024/06/05 03:52

不多说了，都在代码里了，假设只有加法和乘法，除法和减法类似

#include <iostream>#include <stack>using namespace std;bool isNotOperator(char c) {  return c != '*' && c != '+'&& c!= '(' && c != ')';}//获取符号的优先级int getPriority(char c) {  int priority = -1;  switch (c) {    case '+':     priority = 1;     break;    case '*':     priority = 2;     break;    case '(':     priority = 3;     break;    default:     break;  }  return priority;}int main() {  char *s = "a+b*c+(d*e+f)*g";  stack stk;  while (*s != '\0' || !stk.empty()) {    char current = *s;    //如果字符串完毕，只有栈里面有元素，那么就全部弹出并输出    if (current == '\0') {       while(!stk.empty()) {         cout << stk.top();       stk.pop();     } break;    }    // 说明不是操作符,直接输出    if(isNotOperator(current)) {      cout<< current;    } else if (current ==')') {     //1.弹出元素并输出直到'('     while(stk.top() != '(') {       cout << stk.top();       stk.pop();     }     //2.将"("也出栈     stk.pop();    } else {      //下面是普通操作符的情况      //1.弹出栈元素直到发现优先级更低的元素     while(!stk.empty() && getPriority(current) <=getPriority(stk.top()) && stk.top() != '(') {cout << stk.top();stk.pop();     }     //将本操作符入栈     stk.push(current);    }    s++;  }  cout << endl;  return 0;}