leetcode165---Compare Version Numbers

问题描述：

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 < 13.37

就是比较“版本号”大小，给定的版本号version1和version2是字符串类型的，当version1>version2的时候，返回1，小于返回-1，相等返回0。

问题求解：

先分别将version1、version2字符串按’.’分割成多个子串，从前往后，前面的子串大小决定整个串的大小，即如果相等位置子串不等，则当version1>version2的时候，返回1，小于返回-1；如果全部子串都相等，则返回0.

class Solution {public:    int compareVersion(string version1, string version2) {        int v1=0, v2=0;        int n1=version1.size(), n2=version2.size();        while(v1<n1 || v2<n2)        {            string str1="";            string str2="";            while(version1[v1] != '.' && v1<n1)            {//按.截取字符串1的每一段                str1 += version1[v1];                ++v1;            }            while(version2[v2] != '.' && v2<n2)            {//按.截取字符串2的每一段                str2 += version2[v2];                ++v2;            }            if(str1 == "") str1="0";//处理空的情况，对应大小为0            if(str2 == "") str2="0";            int i1=stoi(str1), i2=stoi(str2);            if(i1 != i2)            {                return i1>i2?1:-1;            }            //i1==i2的情况，将两字符串索引后移，判断后面情况            ++v1;            ++v2;        }        return 0;    }};