标签:遍历 ++ 字符串 operator cto token blog discuss atoi
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
1 int evalRPN(vector<string>& tokens) 2 {//字符串可以直接比较。。不用strcmp 直接tokens[i]=="*" 3 stack<int> st; 4 for (int i = 0; i < tokens.size();i++) 5 { 6 if (atoi(tokens[i].c_str())!=0)st.push(atoi(tokens[i].c_str()));//返回非0代表为数字 7 else if (strcmp(tokens[i].c_str(), "0") == 0) 8 { 9 st.push(atoi(tokens[i].c_str())); 10 } 11 else 12 { 13 int b = st.top(); 14 st.pop(); 15 int a = st.top(); 16 st.pop(); 17 if (strcmp(tokens[i].c_str(), "+")==0) st.push(a + b); 18 else if (strcmp(tokens[i].c_str(), "-") == 0)st.push(a - b); 19 else if (strcmp(tokens[i].c_str(), "*") == 0)st.push(a * b); 20 else if (tokens[i] == "/")st.push(a / b); 21 else 22 { 23 //说明有别的字符 错误 24 } 25 } 26 } 27 return st.top(); 28 }
1 int evalRPN(vector<string> &tokens) { 2 stack<int> st; 3 int s1,s2; 4 s1=s2=0; 5 int res=0; 6 for(vector<string>::iterator iter=tokens.begin();iter!=tokens.end();iter++) 7 { 8 if (*iter == "+") 9 { 10 s1=st.top(); 11 st.pop(); 12 s2=st.top(); 13 st.pop(); 14 res=s1+s2; 15 st.push(res); 16 } 17 18 else if (*iter == "-") 19 { 20 s1=st.top(); 21 st.pop(); 22 s2=st.top(); 23 st.pop(); 24 res=s2-s1; 25 st.push(res); 26 } 27 else if (*iter == "*") 28 { 29 s1=st.top(); 30 st.pop(); 31 s2=st.top(); 32 st.pop(); 33 res=s1*s2; 34 st.push(res); 35 } 36 else if (*iter== "/") 37 { 38 s1=st.top(); 39 st.pop(); 40 s2=st.top(); 41 st.pop(); 42 res=s2/s1; 43 st.push(res); 44 } 45 else 46 { 47 st.push(atoi((*iter).c_str())); 48 } 49 } 50 return st.top(); 51 53 }
[leetcode-150-Evaluate Reverse Polish Notation]
标签:遍历 ++ 字符串 operator cto token blog discuss atoi
原文地址:http://www.cnblogs.com/hellowooorld/p/6432921.html