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