标签：blog http io for ar div amp log
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
1st （ 9 tries ）
class Solution
{
public:
bool isnum(string &s)
{
bool minus = false;
if(s[0] == ‘-‘)
minus = true;
if(minus)
{
if(s.length() == 1)
return false;
else
{
for(int i = 1;i < s.length();i++)
{
if(s[i] < ‘0‘ || s[i] > ‘9‘)
return false;
}
return true;
}
}
else
{
for(int i = 0;i < s.length();i++)
{
if(s[i] < ‘0‘ || s[i] > ‘9‘)
return false;
}
return true;
}
}
int evalRPN(vector<string> &tokens)
{
stack<string> numst;
for(int i = 0;i < tokens.size();i++)
{
if(isnum(tokens[i]))
{
numst.push(tokens[i]);
}
else
{
int r = atoi(numst.top().c_str());
numst.pop();
int l = atoi(numst.top().c_str());
numst.pop();
int re = 0;
if(tokens[i] == "+")
{
re = l+r;
}
if(tokens[i] == "-")
{
re = l-r;
}
if(tokens[i] == "*")
{
re = l*r;
}
if(tokens[i] == "/")
{
re = l/r;
}
stringstream ss;
ss<<re;
numst.push(ss.str());
}
}
int re = atoi(numst.top().c_str());
return re;
}
};
2nd （1 tries）
class Solution {
public:
int evalRPN(vector<string> &tokens) {
stack<int> st;
for(int i = 0;i < tokens.size();i++) {
if(tokens[i] == "+") {
int right = st.top();
st.pop();
int left = st.top();
st.pop();
st.push(left+right);
}
else if(tokens[i] == "-") {
int right = st.top();
st.pop();
int left = st.top();
st.pop();
st.push(left-right);
}
else if(tokens[i] == "*") {
int right = st.top();
st.pop();
int left = st.top();
st.pop();
st.push(left*right);
}
else if(tokens[i] == "/") {
int right = st.top();
st.pop();
int left = st.top();
st.pop();
st.push(left/right);
}
else {
int number = atoi( tokens[i].c_str() );
st.push(number);
}
}
return st.top();
}
};
【Leetcode】Evaluate Reverse Polish Notation,布布扣,bubuko.com
【Leetcode】Evaluate Reverse Polish Notation
标签：blog http io for ar div amp log
原文地址：http://www.cnblogs.com/weixliu/p/3924366.html
