leetcode 165

才一周没刷leetcode，手就生了，这个题目不难，但是完全AC还是挺费劲的。

题目描述：

Compare two version numbers version1 and version2.
If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.

You may assume that the version strings are non-empty and contain only digits and the . character.
The . character does not represent a decimal point and is used to separate number sequences.
For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.

Here is an example of version numbers ordering：

0.1 < 1.1 < 1.2 < 12.21

题目不难，就是比较字符串。但是有几点需要注意。

1.各大博客上po出的代码现在都无法AC，原因在于测试case修正了。比如一些算法直接将自连个子版本号分割后对齐转化为int型，现在有个case直接导致这里发生溢出。

2.可能有多个子版本号，这里要考虑全面。

代码如下：

class Solution {
public:
    int compareVersion(string version1, string version2) {
        
        string v1_front, v1_back, v2_front, v2_back;
    
        vector<string> v1;
        vector<string> v2;
        
        splitString(version1, v1);
        splitString(version2, v2);
        
        vector<unsigned long long> v1_long;
        vector<unsigned long long> v2_long;
        
        for (auto i = v1.begin(); i != v1.end(); i ++) {
            if (i->size() != 0)
                v1_long.push_back(stoull(*i));
            else
                v1_long.push_back(0);
            
        }
        
        for (auto i = v2.begin(); i != v2.end(); i ++) {
            if (i->size() != 0)
                v2_long.push_back(stoull(*i));
            else
                v2_long.push_back(0);
            
        }
        
        while (v1_long.size() < v2_long.size()) {
            v1_long.push_back(0);
        }
        
        while (v1_long.size() > v2_long.size()) {
            v2_long.push_back(0);
        }
        
        for (int i = 0; i < v1_long.size(); i ++) {
            if (v1_long[i] != v2_long[i]) {
                return v1_long[i] > v2_long[i] ? 1 : -1;
            }
        }
        return 0;

        
    }
    
    void abandonFrontZeros(string & s)
    {
        int pos = 0;
        while (s[pos] == '0' && pos < s.size()) {
            ++pos;
        }
        s = s.substr(pos, s.size() - pos);
    }

    void splitString(string & s, vector<string> & v)
    {
        vector<int> dot_pos;
        for (int i = 0; i < s.size(); i ++) {
            if (s[i] == '.') {
                dot_pos.push_back(i);
            }
        }
        
        int b = 0;
        for (int i = 0; i < dot_pos.size(); i ++) {
            v.push_back(s.substr(b, dot_pos[i] - b));
            b = dot_pos[i] + 1;
        }
        
        v.push_back(s.substr(b, s.size() - b));
        
        for (auto i = v.begin(); i != v.end(); i ++) {
            abandonFrontZeros(*i);
        }
    }

    
};

感慨一些：C++11的一些新特性还是很方便的。string类的自带接口如substr（）很方便。还有stoi（）， stoull（）， stod（）等模板函数也很方便。