[LeetCode] Compare Version Numbers
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
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
String
分析:将字符串中的每个数字提取出来,保存到数组中,然后逐个比较数组各个元素。
注意:“1” 和 “1.0” 以及“1.0.0” 是相同的,需要单独处理一下。。
也快参考 http://www.cnblogs.com/grandyang/p/4244123.html
class Solution { private: int string2Int(const string& str) { if(str.size() == 0) return 0; int rtn = 0; for(int i = 0; i < str.size(); i++) { rtn = rtn*10 + (str[i] - '0'); } return rtn; } vector<int> string2Ints(const string& str) { vector<int> rtn; if(str.size() == 0) return rtn; int size = str.size(); int left = 0; int tmp; for(int i = 0; i < size; i++) { if(str[i] == '.') { tmp = string2Int(str.substr(left, i - left )); rtn.push_back(tmp); left = i + 1; } } tmp = string2Int(str.substr(left, size + 1 - left )); rtn.push_back(tmp); return rtn; } public: int compareVersion(string version1, string version2) { if(version1 == version2) return false; vector<int> rtn1 = string2Ints(version1); vector<int> rtn2 = string2Ints(version2); //printVector(rtn1); //printVector(rtn2); int size = min(rtn1.size(), rtn2.size()); for(int i = 0; i < size; i++) { if(rtn1[i] > rtn2[i] ) return 1; else if(rtn1[i] < rtn2[i] ) return -1; } if(rtn1.size() == rtn2.size()) return 0; else if(rtn1.size() > rtn2.size()) { for(int i = size; i < rtn1.size(); i++) { if(rtn1[i] != 0) return 1; } return 0;//all is zero } else //if(rtn1.size() < rtn2.size()) { for(int i = size; i < rtn2.size(); i++) { if(rtn2[i] != 0) return -1; } return 0;//all is zero } } };