字符串表达式求值(支持多种类型运算符)
一、说明
1. 输入字符串为中缀表达式,无需转为后缀表达式
2. 支持的运算符包括:
算术运算符:"+,-,*,/"
关系运算符:">,<,>=,<=,=,!="(注意等于运算符采用的是一个等号)
逻辑运算符:"&&,||"
3. 支持大于10的数字,不支持负数操作数,但支持中间结果和返回值为负数
二、算法原理&步骤
本文算法对中缀表达式形式字符串进行求值,同时支持与或运算和逻辑运算(若含有关系运算符或者逻辑运算符,则输出为1或者0)。类似于加减乘除,将关系运算符和逻辑运算符看作优先级低的运算符进行处理,优先级:算术运算符>关系运算符>逻辑运算符。
步骤:
1. 初始化两个空堆栈,一个存放操作数,一个存放运算符。
2. 从左至右扫描输入字符串,依次读取。
- 2.1 若为操作数,则压入操作数栈;
- 2.2 若为运算符,判断其优先级是否大于运算符栈栈顶元素优先级。若大于栈顶元素优先级,则直接压栈;否则,弹出栈顶元素operator,同时依次从操作数栈中弹出两个元素number1,number2,计算表达式(number2 operator number1)的值value,并将值value压入操作数栈。重复上述过程直至当前扫描的操作符优先级大于栈顶元素,然后将当前运算符压栈。
3. 弹出运算符栈顶元素operator,同时依次从操作数栈中弹出两个元素number1,number2,计算表达式(number2 operator number1)的值value,并将值value压入操作数栈。重复上述过程直至运算符栈为空。
4. 此时操作数栈应该只有一个元素,即为表达式的值。
三、代码&测试
求值函数:
1 /* 字符串表达式求值 2 * @param input: 输入的字符串 3 * @param output: 表达式的值,若含有关系运算符则为1或者0 4 * return 计算过程是否正常 5 */ 6 bool ExpValue(string input,int& output) 7 { 8 stack<int> operand_stack; 9 stack<string> operator_stack; 10 11 char prev = 0; // 上一个属于运算符的字符 12 for (int i = 0; i < input.size(); i++) 13 { 14 char c = input[i]; 15 // prev是否是一个完整运算符 16 if (!isOperator(c) && prev) 17 { 18 string new_op = string("").append(1, prev); 19 addNewOperator(new_op, operand_stack, operator_stack); 20 prev = 0; 21 } 22 23 // 数字 24 if (isdigit(c)) 25 { 26 int val_c = c - '0'; 27 if (i > 0 && isdigit(input[i - 1])) 28 { 29 int top_num = operand_stack.top(); 30 top_num = top_num * 10 + val_c; 31 operand_stack.pop(); 32 operand_stack.push(top_num); 33 } 34 else 35 operand_stack.push(val_c); 36 } 37 // 运算符字符 38 else if (isOperator(c)) 39 { 40 // 处理两字符运算符 41 if (prev) 42 { 43 string new_op = string("").append(1, prev).append(1, c); 44 addNewOperator(new_op, operand_stack, operator_stack); 45 prev = 0; 46 } 47 else 48 prev = c; 49 } 50 else if (c == '(') 51 operator_stack.push("("); 52 else if (c == ')') 53 { 54 // 处理括号内的运算符 55 while (operator_stack.top()!="(") 56 { 57 int num1 = operand_stack.top(); 58 operand_stack.pop(); 59 int num2 = operand_stack.top(); 60 operand_stack.pop(); 61 string op = operator_stack.top(); 62 operator_stack.pop(); 63 64 int val = Calculate(num2, num1, op); 65 operand_stack.push(val); 66 } 67 operator_stack.pop(); // 弹出"(" 68 } 69 } 70 assert(operand_stack.size() == operator_stack.size() + 1); 71 // 弹出所有运算符 72 while(!operator_stack.empty()) 73 { 74 int num2 = operand_stack.top(); 75 operand_stack.pop(); 76 int num1 = operand_stack.top(); 77 operand_stack.pop(); 78 string op = operator_stack.top(); 79 operator_stack.pop(); 80 81 int val = Calculate(num1, num2, op); 82 operand_stack.push(val); 83 } 84 85 if (operand_stack.size() == 1) { 86 output = operand_stack.top(); 87 return true; 88 } 89 return false; 90 }
其中用到的子函数有:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
/* 判断字符是否属于运算符 */ bool isOperator(char c) { switch (c) { case '-': case '+': case '*': case '/': case '%': case '<': case '>': case '=': case '!': case '&': case '|': return true; default: return false; } } /* 获取运算符优先级 */ int getPriority(string op) { int temp = 0; if (op == "*" || op == "/" || op == "%") temp = 4; else if (op == "+" || op == "-") temp = 3; else if (op == ">" || op == "<" || op == ">=" || op == "<=" || op == "=" || op == "!=") temp = 2; else if (op == "&&" || op == "||") temp = 1; return temp; } /* * 返回一个两元中缀表达式的值 * syntax: num_front op num_back * @param num_front: 前操作数 * @param num_back: 后操作数 * @param op: 运算符 */ int Calculate(int num_front, int num_back, string op) { if (op == "+") return num_front + num_back; else if (op == "-") return num_front - num_back; else if (op == "*") return num_front * num_back; else if (op == "/") return num_front / num_back; else if (op == "%") return num_front % num_back; else if (op == "!=") return num_front != num_back; else if (op == ">=") return num_front >= num_back; else if (op == "<=") return num_front <= num_back; else if (op == "=") return num_front == num_back; else if (op == ">") return num_front > num_back; else if (op == "<") return num_front < num_back; else if (op == "&&") return num_front && num_back; else if (op == "||") return num_front || num_back; return 0; } /* 新运算符入栈操作 */ void addNewOperator(string new_op, stack<int>& operand_stack, stack<string>& operator_stack) { while (!operator_stack.empty() && getPriority(operator_stack.top()) >= getPriority(new_op)) { int num2 = operand_stack.top(); operand_stack.pop(); int num1 = operand_stack.top(); operand_stack.pop(); string op = operator_stack.top(); operator_stack.pop(); int val = Calculate(num1, num2, op); operand_stack.push(val); } operator_stack.push(new_op); }
测试结果:
int main() { string s0 = "10-1*10+3%2"; string s1 = "100 + (3-33)*2"; string s2 = "20+1 >= 20 && 20+1 < 20"; string s3 = "10>20 || 10/1>=5"; int ret = -1; if (ExpValue(s0, ret)) cout << s0 << "的值: " << ret << endl; if (ExpValue(s1, ret)) cout << s1 << "的值: " << ret << endl; if (ExpValue(s2, ret)) cout << s2 << "的值: " << ret << endl; if (ExpValue(s3, ret)) cout << s3 << "的值: " << ret << endl; return 0; }
上述代码的执行结果为: