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.
Here is an example of version numbers ordering:
0.1 < 1.1 < 1.2 < 13.37
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
将被'.'分割字段逐个取出做比较即可。
class Solution { public: int compareVersion(string version1, string version2) { int len1 = version1.size(); int len2 = version2.size(); int begin1 = 0; int begin2 = 0; int end1 = 0; int end2 = 0; while(end1 != len1 && end2 != len2) { // extract section in version1 string sec1; while(end1 != len1 && version1[end1] != '.') end1 ++; if(end1 == len1) sec1 = version1.substr(begin1); else { sec1 = version1.substr(begin1, end1-begin1); begin1 = end1 + 1; end1 = begin1; } // extract section in version2 string sec2; while(end2 != len2 && version2[end2] != '.') end2 ++; if(end2 == len2) sec2 = version2.substr(begin2); else { sec2 = version2.substr(begin2, end2-begin2); begin2 = end2 + 1; end2 = begin2; } // compare sec1 and sec2 int num1 = atoi(sec1.c_str()); int num2 = atoi(sec2.c_str()); if(num1 > num2) return 1; if(num1 < num2) return -1; } if(end1 == len1 && end2 == len2) return 0; while(end1 != len1) {// version1 remains string sec1; while(end1 != len1 && version1[end1] != '.') end1 ++; if(end1 == len1) sec1 = version1.substr(begin1); else { sec1 = version1.substr(begin1, end1-begin1); begin1 = end1 + 1; end1 = begin1; } int num1 = atoi(sec1.c_str()); if(num1 > 0) return 1; } while(end2 != len2) {// version2 remains string sec2; while(end2 != len2 && version2[end2] != '.') end2 ++; if(end2 == len2) sec2 = version2.substr(begin2); else { sec2 = version2.substr(begin2, end2-begin2); begin2 = end2 + 1; end2 = begin2; } int num2 = atoi(sec2.c_str()); if(num2 > 0) return -1; } return 0; } };