leetcode 字符串处理Compare Version Numbers
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【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
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。
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; } };
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