[LeetCode] Compare Version Numbers

Compare two version numbers version1 and version1.
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

逐位比较就好了，把字符串转化成整型，这样就可以忽略前置0的影响了。

class Solution {
public:
    int compareVersion(string version1, string version2) {
        int val1, val2;
        int idx1 = 0, idx2 = 0;
        while (idx1 < version1.length() || idx2 < version2.length()) {
            val1 = 0; 
            while (idx1 < version1.length()) {
                if (version1[idx1] == '.') {
                    ++idx1;
                    break;
                }
                val1 = val1 * 10 + (version1[idx1] - '0');
                ++idx1;
            }
            val2 = 0; 
            while (idx2 < version2.length()) {
                if (version2[idx2] == '.') {
                    ++idx2;
                    break;
                }
                val2 = val2 * 10 + (version2[idx2] - '0');
                ++idx2;
            }
            if (val1 > val2) return 1;
            if (val1 < val2) return -1;
        }
        return 0;
    }
};