[LeetCode]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

version number,顾名思义就是版本号的意思啦。它题目中给的例子不太直观，我自己举一个例子就是:

1.2 < 1,10

因为1.10和1.1是不一样的，1.10代表the tenth revision,而1.1代表the first revision。所以这道题要将version string按照“.”分段，然后逐段比较。 有一点需要注意的是，一开始我的代码wrong answer，看到它卡在测试案例

“1.1”,”1.01.0”

对两个version string预处理后，得到的两组数据分别为：1,1和1,1,0。其实这种情况是相等的，但是一开始的代码中，数组没有初始化，array1[2]为负数，，小于0，所以返回-1。修改最后的AC代码为：

class Solution {public:vector<string> split(string s1,char c){vector<string> ans;string::size_type pos1, pos2;pos2= s1.find(c);pos1 = 0;while (string::npos != pos2){ans.push_back(s1.substr(pos1,pos2-pos1));pos1 = pos2 + 1;pos2 = s1.find(c, pos1);}if (pos1 != s1.length()){ans.push_back(s1.substr(pos1));}return ans;}int compareVersion(string version1, string version2) {vector<string> s1 = split(version1, ‘.’);vector<string> s2 = split(version2, ‘.’);int maxLen = s1.size()>s2.size() ? s1.size() : s2.size();int* array1 = new int[maxLen];int* array2 = new int[maxLen];memset(array1, 0, maxLen*sizeof(int));memset(array2, 0, maxLen*sizeof(int));int i = 0;for (i = 0; i < s1.size(); i++){array1[i] = atoi(s1[i].c_str());}for (i = 0; i < s2.size(); i++){array2[i] = atoi(s2[i].c_str());}for (i = 0; i < maxLen; i++){if (array1[i]>array2[i])return 1;else if (array1[i] < array2[i])return -1;}return 0;}};

拥有一颗比九万五千公里还辽阔的心，