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
LeetCode Source
分析:写的比较乱,这个题目不难,注意好边界条件。把version转换成了大版本int型m,和小版本float型n。
class Solution { public: int compareVersion(string version1, string version2) { int m1=0; float n1=0; int m2=0; float n2=0; ToInt(version1,m1,n1); ToInt(version2,m2,n2); if(m1>m2) return 1; if(m1<m2) return -1; if(m1==m2){ if(n1>n2) return 1; if(n1<n2) return -1; if(n1==n2) return 0; } } void ToInt(string s,int &m,float &n){ int sz=s.size(); int flag = 0; int i = 0; for(;i<sz;++i){ if(s[i]=='.') break; else{ m = m*10+s[i]-'0'; } } for(i=i+1;i<sz;++i){ if(s[i]=='.') { if(s[i+1]!='0') flag ++; if(s[i+1]=='0'&&i+2>sz-1) break; continue; } n = n*10+s[i]-'0'; } while(flag>0){ n=n/10; --flag; } } };