LeetCode *** 165. Compare Version Numbers

题目：

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

分析：

因为只有数字与'.'所以分开分析即可。

代码：

class Solution {
public:
int compareVersion(string version1, string version2) {

int i=0,j=0;
int ilen=version1.length(),jlen=version2.length();

while(i<ilen&&j<jlen){

unsigned int n1=0;
unsigned int n2=0;

while(version1[i]!='.'&&i<ilen){
n1=n1*10+version1[i]-'0';
i++;
}

while(version2[j]!='.'&&j<jlen){
n2=n2*10+version2[j]-'0';
j++;
}

if(n1>n2)return 1;
if(n1<n2)return -1;

if(i<ilen)++i;
if(j<jlen)++j;

}

while(i<ilen){
unsigned int n1=0;

while(version1[i]!='.'&&i<ilen){
n1=n1*10+version1[i]-'0';
i++;
}
if(i<ilen)++i;
if(n1)return 1;
}

while(j<jlen){
unsigned int n2=0;

while(version2[j]!='.'&&j<jlen){
n2=n2*10+version2[j]-'0';
j++;
}
if(j<jlen)++j;
if(n2)return -1;
}

return 0;

}
};