Compare Version Numbers
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.
考虑以下几种情况即可：
Example1: version1=="11.22.33", version2=="11.22.22". 11 == 11; 22 == 22; 33 > 22; return 1.
Example2: version1=="11.22.33", version2=="11.22.33". 11 == 11; 22 == 22; 33 == 33; return 0.
Example3: version1=="11.22.33", version2=="11.22.33.00.00". 11 == 11; 22 == 22; 33 == 33; remaining version2 equals to 0; return 0.
Example4: version1=="11.22.33.00.01", version2=="11.22.33". 11 == 11; 22 == 22; 33 == 33; remaining version1 contains 01; return 1.
第一种方法，把字符串拆分到vector中，然后进行对比：
1 class Solution {
2 public:
3
4
5 void split(std::string& s, std::string& delim,std::vector< std::string >& ret)
6 {
7 size_t last = 0;
8 size_t index=s.find_first_of(delim,last);
9 while (index!=std::string::npos)
10 {
11 ret.push_back(s.substr(last,index-last));
12 last=index+1;
13 index=s.find_first_of(delim,last);
14 }
15 if (index-last>0)
16 {
17 ret.push_back(s.substr(last,index-last));
18 }
19 }
20
21 int compareVersion(string version1, string version2) {
22
23 //拆分到vector中
24 vector<string> v1,v2;
25 string delim=".";
26
27 split(version1,delim,v1);
28 split(version2,delim,v2);
29
30 int n=v1.size()>v2.size()?v1.size():v2.size();
31
32
33 for(int i=0;i<n;i++)
34 {
35
36 //当v1完了，只需要看v2末尾是否都为0即可
37 if(i>=v1.size())
38 {
39 if(atoi(v2[i].c_str())!=0)
40 {
41 return -1;
42 }
43 }
44
45 //当v2完了，只需要看v1末尾是否都为0即可
46 if(i>=v2.size())
47 {
48 if(atoi(v1[i].c_str())!=0)
49 {
50 return 1;
51 }
52 }
53
54 //v1,v2没完，则比较大小即可
55 if(i<v1.size()&&i<v2.size())
56 {
57 if(atoi(v1[i].c_str())>atoi(v2[i].c_str()))
58 {
59 return 1;
60 }
61 if(atoi(v1[i].c_str())<atoi(v2[i].c_str()))
62 {
63 return -1;
64 }
65 }
66 }
67 return 0;
68 }
69 };
第二种解法，采用库函数
C++ 11中，stoi可以从字符串string中取得数字，并返回不是数字下一个位置。
string中的erase函数：string& erase (size_t pos = 0, size_t len = npos)
1 class Solution {
2 public:
3
4 int compareVersion(string version1, string version2) {
5
6
7 while(!(version1.empty()&&version2.empty()))
8 {
9 size_t pos1=0;
10 size_t pos2=0;
11 int n1,n2;
12
13 n1=version1.empty()?0:stoi(version1,&pos1);
14 n2=version2.empty()?0:stoi(version2,&pos2);
15
16
17 if(n1>n2)
18 return 1;
19 else if(n1<n2)
20 return -1;
21
22
23 if(!version1.empty())
24 version1.erase(0,pos1+1);
25 if(!version2.empty())
26 version2.erase(0,pos2+1);
27 }
28
29 return 0;
30
31 }
32 };