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

分析：题意为 比较两个版本数字version1和version2，值得注意的是.并不是代表小数点，例如：2.5表示第二个一级第五个二级版本

思路：可以进行逐位比较，把字符串转化成整型。

注意考虑test case：0.1 与 0.01  结果是：0.1>0.01

代码如下：

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class Solution {   public:       int compareVersion(string version1, string version2) {          int v1, v2;           int t1 = 0, t2 = 0;           while (t1 < version1.length() || t2 < version2.length()) {               v1 = 0;               while (t1 < version1.length()) {                   if (version1[t1] == '.') {                      ++t1;                      break;                  }                  v1 = v1 * 10 + (version1[t1] - '0');                  ++t1;              }              v2 = 0;              while (t2 < version2.length()) {                  if (version2[t2] == '.') {                      ++t2;                      break;                  }                  v2 = v2 * 10 + (version2[t2] - '0');                  ++t2;             }              if (v1 > v2) return 1;              if (v1 < v2) return -1;          }          return 0;      }  };

　其他解法：

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class Solution {   public:       int compareVersion(string version1, string version2) {         const char *p1 = version1.c_str()-1;         const char *p2 = version2.c_str()-1;         do{             int v1 = 0, v2 =0;             if (p1){                 v1=atoi(p1+1);                 p1 = strchr(p1+1, '.');             }             if (p2){                 v2=atoi(p2+1);                 p2 = strchr(p2+1, '.');             }             if (v1<v2) return -1;             if (v2<v1) return 1;         }while(p1||p2);         return 0;     }  };

　或：

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class Solution { public:     int compareVersion(string version1, string version2) {         int n = version1.size(), m = version2.size();         for (int i = 0, j = 0; i < n || j < m; i++, j++) {             size_t p = 0;             int a = ((i >= n) ? 0 : stoi(version1.substr(i), &p));             i += p;             int b = ((j >= m) ? 0 : stoi(version2.substr(j), &p));             j += p;             if (a > b) return 1;             if (a < b) return -1;         }         return 0;     } };

　　或：

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class Solution {   public: int compareVersion(string version1, string version2) {     istringstream iss1(version1), iss2(version2);     string token1, token2;       while(!iss1.eof() || !iss2.eof()) {         getline(iss1, token1, '.'); getline(iss2, token2, '.');           if(stoi(token1) > stoi(token2)) return 1;         if(stoi(token1) < stoi(token2)) return -1;           token1 = token2 = "0";     }       return 0; } };