[Leetcode] 165. Compare Version Numbers 解题报告

题目：

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：

我刚刚开始的时候，想当然的以为version字符串中最多只会出现一个“.”，结果就悲剧了。看来在开始coding之前和面试官做clarification是多么多么的重要！思路是：每次获取位于“.”之间的子串，然后把它转换成为整数进行比较。一旦比较出来结果，就直接返回，否则开始比较下一个子串。不过在while循环结束后需要当心，如果一个字符串还没有被处理完，需要判断其剩余部分的版本号是否大于0。具体请见代码片段中的最后三行。

代码：

class Solution {public:    int compareVersion(string version1, string version2) {        version1 += ".", version2 += ".";        int i = 0, j = 0;        int length1 = version1.size(), length2 = version2.size();        while (i < length1 && j < length2) {            int pos1 = version1.find('.', i), pos2 = version2.find('.', j);            string tem1 = version1.substr(i, pos1 - i);            string tem2 = version2.substr(j, pos2 - j);            int val1 = stoi(tem1), val2 = stoi(tem2);            if (val1 > val2) {                return 1;            }            if (val1 < val2) {                return -1;            }            i = pos1 + 1, j = pos2 + 1;        }        if (i < length1 && stof(version1.substr(i)) > 0) {      // check the remaining part            return 1;        }        if (j < length2 && stof(version2.substr(j)) > 0) {      // check the remaining part            return -1;        }        return 0;    }};