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.

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本题要求比较两个版本号的大小。简单题，按点.分隔，然后转换为int依次比较就好。 为了防止有的版本号没有点.，开始先给两个版本号末尾添加上一个点。另外1.0和1这两个版本号是相等的，所以如果版本位数有差别时，需要补0，而不是简单的认为长的比短的大。 完整代码如下：

class Solution {
public:
    int compareVersion(string version1, string version2)
    {
        version1 += ".";
        version2 += ".";
        size_t p1 = version1.find(‘.’), p2 = version2.find(‘.’);
        int v1, v2;
        while (p1 != string::npos || p2 != string::npos) {
            if (p1 == string::npos)
                v1 = 0;
            else {
                v1 = atoi(version1.substr(0, p1).c_str());
                version1 = version1.substr(p1 + 1);
                p1 = version1.find(‘.’);
            }
            if (p2 == string::npos)
                v2 = 0;
            else {
                v2 = atoi(version2.substr(0, p2).c_str());
                version2 = version2.substr(p2 + 1);
                p2 = version2.find(‘.’);
            }
            if (v1 < v2)
                return -1;
            else if (v1 > v2)
                return 1;
        }
        return 0;
    }
};

本代码提交AC，用时3MS。

二刷。更简洁的代码：

class Solution {
private:
	void ParseVersion(const string &version, vector<int> &v) {
		int n = version.size();
		int i = 0, j = 0;
		while (i < n) {
			j = i + 1;
			while (j < n&&version[j] != '.')++j;
			int tmp = atoi(version.substr(i, j - i).c_str());
			v.push_back(tmp);
			i = j + 1;
		}
	}
public:
	int compareVersion(string version1, string version2) {
		vector<int> v1, v2;
		ParseVersion(version1, v1);
		ParseVersion(version2, v2);
		int i = 0, j = 0, n1 = v1.size(), n2 = v2.size();
		while (i < n1 || j < n2) {
			int a = (i < n1 ? v1[i] : 0);
			int b = (j < n2 ? v2[j] : 0);
			if (a > b)return 1;
			else if (a < b)return -1;
			else {
				++i;
				++j;
			}
		}
		return 0;
	}
};

本代码提交AC，用时8MS。