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：

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：

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：

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：

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;
}
};