【leetcode 字符串处理】Compare Version Numbers
leetcode新题,Compare Version Numbers,字符串处理。题意:题意很清晰,就是比较“版本号”大小,给定的版本号version1和version2是字符串类型的,当version1>version2的时候,返回1,反之返回-1。这道题属于细节处理题,除了字符串处理繁琐一点之外没有什么。解题思路:先分别将version1、version2字符串按'.'分割成多个子串,每个子串
【leetcode 字符串处理】Compare Version Numbers
1、题目
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
2、分析
3、代码
class Solution {
public:
int compareVersion(string version1, string version2) {
vector<int> result1=getInt(version1);
vector<int> result2=getInt(version2);
int len1=result1.size();
int len2=result2.size();
if(len2<len1) return -1*compareVersion(version2, version1);
int i=0;
while(i<len1 && result1[i]==result2[i]) i++;
if(i==len1){ //str1和str2前len1位都相等,则看看str2后面的len2-len1位是否都为0即可判断它们的大小
int j=len2-1;
while(j >= len1){
if(result2[j--]!=0) return -1;
}
return 0;
}else{ //str1和str2前len1位不都相等,直接判断第i位
if(result1[i]<result2[i]) return -1;
else return 1;
}
}
private:
//将version字符串按'.'拆成多个,转化为整型放入容器
vector<int> getInt(string version){
vector<int> result;
int len=version.size();
int pre=0;
for(int i=0;i<len;i++){
if(version[i]=='.'){
string str(version.begin()+pre,version.begin()+i); //注意这种初始化形式,左闭右开,即str不包括version[version.begin()+i]
result.push_back(stoi(str));
pre=i+1;
}
}
string str(version.begin()+pre,version.end());
result.push_back(stoi(str));
return result;
}
};
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