leetcode 165. 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.

You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.

Example 1:

Input: version1 = "0.1", version2 = "1.1"
Output: -1

Example 2:

Input: version1 = "1.0.1", version2 = "1"
Output: 1

Example 3:

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

Example 4:

Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”

Example 5:

Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"

Note:

1. Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.
2. Version strings do not start or end with dots, and they will not be two consecutive dots.

题目归类：简单题

题目大意：比较版本数的大小。

思路：硬编码。比较相同位置的数值大小

C++代码：

class Solution {
public:
    int compareVersion(string version1, string version2) {
        int flag = 0;
        int s1 = 0, s2 = 0;
        int len1 = version1.length(), len2 = version2.length();
        while (s1 < len1 || s2 < len2) {
            int n1 = 0, n2 = 0;
            while ((s1 < len1) && (version1[s1] != '.')) {
                n1 = n1 * 10 + (version1[s1] - '0');
                ++s1;
            }
            while ((s2 < len2) && (version2[s2] != '.')) {
                n2 = n2 * 10 + (version2[s2] - '0');
                ++s2;
            }
            if (n1 > n2) {
                flag = 1;
                break;
            } 
            else if (n1 < n2) {
                flag = -1;
                break;
            } 
            ++s1;
            ++s2;
        }
        return flag;
    }
};