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,
- .
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.
- 2.5
Here is an example of version numbers ordering:
- 0.1 < 1.1 < 1.2 < 13.37
最后再比較两个容器中相应位置的数的大小。
当然须要考虑它们长度不同的情况。
注意点:(1)version 的形式:10.23.1、01.23、1.2.3.4.... 诸如此类(2)字符串转化成整型,我的程序中直接用库函数 stoi()
- 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就可以推断它们的大小
- 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。直接推断第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); //注意这样的初始化形式,左闭右开,即str不包含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;
- }
- };
【leetcode 字符串处理】Compare Version Numbers
来源: http://www.bubuko.com/infodetail-2116468.html