15
04/2015
[LeetCode] Evaluate Reverse Polish Notation
Evaluate Reverse Polish Notation
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
解题思路：
后缀表达式计算的问题，做过计算机的同学应该都知道。遇到数字，进栈；遇到操作符，将前面两个操作数出栈，计算，将结果入栈。最后的结果会在栈中。
代码如下：
class Solution {
public:
int evalRPN(vector<string>& tokens) {
stack<int> s;
int len = tokens.size();
int num1, num2;
for(int i = 0; i<len; i++){
if(tokens[i]=="+"){
num1 = s.top();
s.pop();
num2 = s.top();
s.pop();
num1 = num2 + num1;
s.push(num1);
}else if(tokens[i] == "-"){
num1 = s.top();
s.pop();
num2 = s.top();
s.pop();
num1 = num2 - num1;
s.push(num1);
}else if(tokens[i] == "*"){
num1 = s.top();
s.pop();
num2 = s.top();
s.pop();
num1 = num2 * num1;
s.push(num1);
}else if(tokens[i] == "/"){
num1 = s.top();
s.pop();
num2 = s.top();
s.pop();
num1 = num2 / num1;
s.push(num1);
}else{
s.push(stringToInt(tokens[i]));
}
}
return s.top();
}
int stringToInt(const string& s){
int len = s.length();
int result = 0;
int sign = 1;
int i = 0;
if(s[0]=='-'){
sign = -1;
i=1;
}
while(i<len){
result = s[i] - '0' + result * 10;
i++;
}
return sign * result;
}
};
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