[leetcode] 165. Compare Version Numbers 解题报告

题目链接：https://leetcode.com/problems/compare-version-numbers/

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 < 13.37

思路：五六个月前尝试过，WA到死，搞得我对这题都有心理阴影了。让我想起大一的时候第一次参加ACM小组做poj1001大数运算的时候，各种恶心的测试数据。

会有12.13.19等等可以连续出现版本，所以一个比较明确的思路是，将字符串以‘.’分割成整数，然后从前往后比较，一旦发现不一样的就说明可以判定大小了。

另外还有就是会出现比如10.10.10和10.10.00这种数据。

代码如下：

class Solution {
public:
    void change(vector<int>& vec, string version)//将字符串转为整型
    {
        int i = 0;
        while(i < version.size())
        {
            stringstream ss;
            int val;
            size_t pos = version.find('.', i);
            string str = version.substr(i, pos-i);
            ss << str;
            ss >> val;
            vec.push_back(val);
            if(pos == string::npos) return;//防止溢出
            i = pos+1;
        }
    }
    int compareVersion(string version1, string version2) {
        vector<int> vec1;
        vector<int> vec2;
        change(vec1, version1);
        change(vec2, version2);
        for(int i =0; i < vec1.size() || i < vec2.size(); i++)
        {
            int val1 =0, val2=0;
            if(i < vec1.size()) val1 = vec1[i];
            if(i < vec2.size()) val2 = vec2[i];
            if(val1 > val2) return 1;
            else if(val1 < val2) return -1;
        }
        return 0;
    }
};