[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：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。如果相同位数无法比较大小, 那就看剩下的长度组成的数字是否大于0. 如果仍无法分出大小, 那就是相等了.

代码如下：

class Solution {public:    int compareVersion(string version1, string version2) {        version1 += ".", version2 += ".";        int i=0, j =0, len1 = version1.size(), len2 = version2.size();        while(i < len1 && j < len2)        {            int pos1 = version1.find('.', i), pos2 = version2.find('.', j);            string tem1 = version1.substr(i, pos1-i);            string tem2 = version2.substr(j, pos2-j);            if(stoi(tem1) > stoi(tem2)) return 1;            if(stoi(tem1) < stoi(tem2)) return -1;            i = pos1+1, j = pos2+1;        }        if(i < len1 && stof(version1.substr(i)) > 0) return 1;         if(j < len2 && stof(version2.substr(j)) > 0) return -1;        return 0;    }};