LeetCode-Compare Version Numbers
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Compare two version numbers version1 and version1.
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.
结题报告:题目给的例子,说的不太明白,版本号还有1.2.4 01.2.4.3这种类型的。所以只要搜索'.'。再进行判断就行
class Solution {public: int compareVersion(string version1, string version2) { int temp1 = 0; int temp2 = 0; string::iterator t1 = version1.begin(); string::iterator t2 = version2.begin(); while(t1 != version1.end() || t2 != version2.end()) { while(*t1 != '.' && t1 != version1.end()) { temp1 = temp1 *10 + *t1++ - '0'; } while(*t2 != '.' && t2 != version2.end()) { temp2 = temp2 *10 + *t2++ - '0'; } if(temp1 > temp2) return 1; else if(temp1 < temp2) return -1; else { if(t1 == version1.end() && t2 == version2.end()) return 0; else if(t1 == version1.end() && t2 != version2.end()) { temp1 = 0; temp2 = 0; t2++; } else if(t1 != version1.end() && t2 == version2.end()) { temp1 = 0; temp2 = 0; t1++; } else { temp1 = 0; temp2 = 0; t1++; t2++; } } } return 0; }};
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