Reverse Integer

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

 

此题注意边界条件 -2^31 < int < 2 ^ 31 - 1    在C++里，9 * 1000,000,000 = 410065408， 不等于9,000,000,000. 但是 2 * 1,000,000,000 = 2,000,000,000.

 

 1 class Solution {
 2 public:
 3     int reverse(int x) {
 4         if(x == 0) return 0;
 5         
 6         bool positive = true;
 7         if(x < 0){
 8             positive = false;
 9             x = -x;
10         }
11         vector<int> v;
12         while(x){
13             v.push_back(x % 10);
14             x /= 10;
15         }
16         int n = v.size();
17         long long temp = 0;
18         for(int i = 0; i < n; i++)
19             temp += v[n - 1 - i] * (pow(10, i));
20         if(temp > INT_MAX || (!positive && temp < INT_MIN)) return 0;
21         else 
22             return positive ? temp : -temp;
23     }
24 };

 

 1 class Solution {
 2 public:
 3     int reverse(int x) {
 4         if(x == -2147483648) return 0;
 5         string s;
 6         int positive = 1;
 7         if(x < 0) {x = -x; positive = -1;}
 8         while(x){
 9             char temp = x % 10 + '0';
10             s += temp;
11             x /= 10;
12         }
13         int result = string2int(s);
14         if(result){
15             if(positive == -1) return -result;
16             return result;
17         }
18         return 0;
19     }
20     
21     int string2int(string s){
22         if(s.length() > 10 || (s.length() == 10 && s[0] > '2')) return 0;
23         double result = 0;
24         int exponential = 1;
25         for(int i = s.length() - 1; i >= 0; i--){
26             int temp = s[i] - '0';
27             result += temp * exponential;
28             exponential *= 10;
29         }
30         if(result > 2147483647.0) return 0;
31         return result;
32     }
33 };

 

class Solution {
public:
    int reverse(int x) {
        if(x == 0) return 0;
        
        bool positive = true;
        if(x < 0){
            positive = false;
            x = -x;
        }
        vector<int> v;
        while(x){
            v.push_back(x % 10);
            x /= 10;
        }
        int n = v.size();
        long long temp = 0;
        for(int i = 0; i < n; i++)
            temp += v[n - 1 - i] * (pow(10, i));
        if(temp > INT_MAX || (!positive && temp < INT_MIN)) return 0;
        else 
            return positive ? temp : -temp;
    }
};

 

posted @ 2015-04-12 11:20  amazingzoe  阅读(198)  评论(0)  编辑  收藏  举报