LeetCode: Evaluate Reverse Polish Notation

这题只要理解RPN的原理就不难，用一个stack就行了。为虾米leetcode不提供atoi函数~~只好自己临时写个。注意负数的情况就可以了

class Solution {
public:
    int stringtoint(string s) {
        int ans = 0;
        for (int i = isdigit(s[0])? 0 : 1; i < s.size(); i++) {
            ans = ans*10 + s[i] - '0';
        }
        return isdigit(s[0])? ans : -ans;
    }
    int evalRPN(vector<string> &tokens) {
        int res = 0;
        stack<int> S;
        for (int i = 0; i < tokens.size(); i++) {
            if (isdigit(tokens[i][0]) || tokens[i].size() > 1 && isdigit(tokens[i][1])) S.push(stringtoint(tokens[i]));
            else {
                int a = S.top();
                S.pop();
                int b = S.top();
                S.pop();
                if (tokens[i] == "+") S.push(b+a);
                if (tokens[i] == "-") S.push(b-a);
                if (tokens[i] == "*") S.push(b*a);
                if (tokens[i] == "/") S.push(b/a);
            }
        }
        return S.top();
    }
};

 C#

public class Solution {
    public int EvalRPN(string[] tokens) {
        int ans = 0;
        Stack<int> S = new Stack<int>();
        for (int i = 0; i < tokens.Length; i++) {
            if (isdigit(tokens[i][0]) || tokens[i].Length > 1 && isdigit(tokens[i][1])) S.Push(stringtoint(tokens[i]));
            else {
                int a = S.Peek();
                S.Pop();
                int b = S.Peek();
                S.Pop();
                if (tokens[i] == "+") S.Push(b + a);
                if (tokens[i] == "-") S.Push(b - a);
                if (tokens[i] == "*") S.Push(b * a);
                if (tokens[i] == "/") S.Push(b / a);
            }
        }
        return S.Peek();
    }
    public bool isdigit(char c) {
        return c <= '9' && c >= '0';
    }
    public int stringtoint(string s) {
        int ans = 0;
        for (int i = isdigit(s[0])? 0 : 1; i < s.Length; i++) {
            ans = ans * 10 + s[i] - '0';
        }
        return isdigit(s[0])? ans : -ans;
    }
}