- 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:
- 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.
解 解析出每一段版本号进行比较，假设解析出来的数字分别为n1, n2
- n1 > n2：版本1>版本2
- n1 < n2：版本1 < 版本2
- n1 == n2 ：比较下一段
在比较到末尾时，长的版本号剩余部分如果全部为0，则版本1 == 版本2，否则长的号对应的版本高
为了方便，在每个版本号后面增加了'#'作为结束符
class Solution {
public:
int compareVersion(string version1, string version2) {
version1.push_back('#');
version2.push_back('#');
int pos1 = 0, pos2 = 0;
int num1 = 0, num2 = 0;
while(version1[pos1] != '#' && version2[pos2] != '#'){
num1 = 0; num2 = 0;
if(version1[pos1] == '.')pos1++;
while(version1[pos1] != '#' && version1[pos1] != '.'){
num1 = num1 * 10 + (version1[pos1++] - '0');
}
if(version2[pos2] == '.')pos2++;
while(version2[pos2] != '#' && version2[pos2] != '.'){
num2 = num2 * 10 + (version2[pos2++] - '0');
}
if(num1 < num2)return -1;
else if(num1 > num2)return 1;
}
if(version1[pos1] == '#' && version2[pos2] == '#'){
return 0;
}else if(version1[pos1] == '#'){
if(allZeros(version2.substr(pos2)))return 0;
else return -1;
}else{
if(allZeros(version1.substr(pos1)))return 0;
else return 1;
}
}
bool allZeros(const string &s){
int test = 0, pos = 0;
while(s[pos] != '#'){
if(s[pos] == '.')pos++;
else test += s[pos++] - '0';
}
if(test == 0)return true;
else return false;
}
};