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

 1 class Solution {
 2 public:
 3     int stringtoint(string s) {
 4         int ans = 0;
 5         for (int i = isdigit(s[0])? 0 : 1; i < s.size(); i++) {
 6             ans = ans*10 + s[i] - '0';
 7         }
 8         return isdigit(s[0])? ans : -ans;
 9     }
10     int evalRPN(vector<string> &tokens) {
11         int res = 0;
12         stack<int> S;
13         for (int i = 0; i < tokens.size(); i++) {
14             if (isdigit(tokens[i][0]) || tokens[i].size() > 1 && isdigit(tokens[i][1])) S.push(stringtoint(tokens[i]));
15             else {
16                 int a = S.top();
17                 S.pop();
18                 int b = S.top();
19                 S.pop();
20                 if (tokens[i] == "+") S.push(b+a);
21                 if (tokens[i] == "-") S.push(b-a);
22                 if (tokens[i] == "*") S.push(b*a);
23                 if (tokens[i] == "/") S.push(b/a);
24             }
25         }
26         return S.top();
27     }
28 };

 1 public class Solution {
 2     public int EvalRPN(string[] tokens) {
 3         int ans = 0;
 4         Stack<int> S = new Stack<int>();
 5         for (int i = 0; i < tokens.Length; i++) {
 6             if (isdigit(tokens[i][0]) || tokens[i].Length > 1 && isdigit(tokens[i][1])) S.Push(stringtoint(tokens[i]));
 7             else {
 8                 int a = S.Peek();
 9                 S.Pop();
10                 int b = S.Peek();
11                 S.Pop();
12                 if (tokens[i] == "+") S.Push(b + a);
13                 if (tokens[i] == "-") S.Push(b - a);
14                 if (tokens[i] == "*") S.Push(b * a);
15                 if (tokens[i] == "/") S.Push(b / a);
16             }
17         }
18         return S.Peek();
19     }
20     public bool isdigit(char c) {
21         return c <= '9' && c >= '0';
22     }
23     public int stringtoint(string s) {
24         int ans = 0;
25         for (int i = isdigit(s[0])? 0 : 1; i < s.Length; i++) {
26             ans = ans * 10 + s[i] - '0';
27         }
28         return isdigit(s[0])? ans : -ans;
29     }
30 }

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posted @ 2014-01-08 08:36 ying_vincent 阅读(171) 评论(0) 收藏举报

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