gatieme / CodingInterviews
剑指Offer——名企面试官精讲典型编程题
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Crawler view#include #include #include #include using namespace std; //判断是否为操作符 bool isOperator(char op) { return (op == '+' || op == '-' || op == '*' || op == '/'); } //判断是否为数字 bool isDigit(char op) { return (op >= '0' && op s1;//创建辅助栈 int left = 0, right = 0;//定义左/右操作数 int result = 0;//定义中间结果 string temp; for (unsigned int i = 0; i 0) //防止运算符后面跟分隔符，所以判断一下temp里面是否有数字 { s1.push(atoi(temp.c_str())); temp.clear(); } } else if (isOperator(tokens[i]) && !s1.empty()) { //防止数字后面直接跟运算符，所以这里也要判断一下temp是否还有数字没有提取出来 if (temp.size() > 0) { s1.push(atoi(temp.c_str())); temp.clear(); } right = s1.top(); s1.pop(); left = s1.top(); s1.pop(); result = cal(left, right, tokens[i]); s1.push(result); } } if (!s1.empty()) { result = s1.top(); s1.pop(); } return result; } //中缀表达式转逆波兰表达式 vector infixToPRN(const string & token) { vector v1; int len = token.size();//string的长度 if (len == 0) return v1; stack s1;//存放逆波兰式的结果 int outLen = 0; for (int i = 0; i pirority(s1.top())) { s1.push(token[i]);// 当前操作符优先级比栈顶高， 将栈顶操作符写入后缀表达式 break; }else { v1.push_back(s1.top()); s1.pop(); } } } while (!s1.empty())//输入结束，将栈中剩余操作符出栈输出 { v1.push_back(s1.top()); s1.pop(); } return v1; } int main() { //vector sRPN = {"4", "13", "5" , "/", "+"};//逆波兰表达式 //cout fix = {"4", "+", "13", "/", "5"};//中缀表达式 string fix = "2+2*(1*2-4/2)*1"; vector RPN = infixToPRN(fix); string s_fix; for (auto it = RPN.begin(); it != RPN.end(); it++) { cout << *it << " "; s_fix += *it; s_fix += " "; } cout << endl; cout << "逆波兰表达式结果为：" << evalRPN(s_fix) << endl; system("pause"); return 0; }