leetcode-Evaluate Reverse Polish Notation

Question:

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Some examples:

[“2”, “1”, “+”, “3”, ““] -> ((2 + 1) 3) -> 9

[“4”, “13”, “5”, “/”, “+”] -> (4 + (13 / 5)) -> 6

Solution:

class Solution {
public:
int evalRPN(vector<string>& tokens) {
int len = tokens.size();
int stack[5000];
int top = -1;
int t1 = 0;
int t2 = 0;
for(int i = 0 ; i < len ; i++){
if(top >= 1){
t1 = stack[top];
t2 = stack[top-1];
}
switch(tokens[i][0]){
case '+':stack[--top]  = t1+t2;break;
case '*':stack[--top]  = t2*t1;break;
case '/':stack[--top]  = t2/t1;break;
case '-':if(tokens[i].size()==1){
stack[--top]  = t2-t1; break;
}
default: stack[++top] = atoi(tokens[i].c_str());
//stringstream ss;
//         ss<<tokens[i];
//         ss>>stack[++top];

}
}
return stack[0];
}
};

总结：

1、后缀表达式求值，典型的栈应用。相应的还有中缀表达式转换后缀表达式

：使用两个栈，一个存储运算符号，一个存储数字，运算符栈一直要保证栈顶运算符号优先级最大。

2、stringstream最是耗时beat 8%，16ms

改用atoi beat 72%，9ms。效率相差一倍