LeetCode: 165. Compare Version Numbers
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LeetCode: 165. Compare Version Numbers
题目描述
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.
Example 1:
Input: version1 = "0.1", version2 = "1.1"
Output: -1
Example 2:
Input: version1 = "1.0.1", version2 = "1"
Output: 1
Example 3:
Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1
解题思路
模拟,依次比较每个版本号。
AC 代码
class Solution {
vector<int> splitVersion(string version)
{
vector<int> verNums;
int beg = 0, end = 0;
while((end = version.find_first_of('.', beg)) != version.npos || beg < version.size())
{
if(end == version.npos) end = version.size();
string subVersion = version.substr(beg, end-beg);
verNums.push_back(stoi(subVersion));
beg = end+1;
}
return verNums;
}
int compareVersion(const vector<int>& ver1, const vector<int>& ver2, int idx)
{
if(idx >= ver1.size() && idx >= ver2.size()) return 0;
else if(idx >= ver1.size() && ver2[idx] != 0) return -1;
else if(idx >= ver1.size() && ver2[idx] == 0) return compareVersion(ver1, ver2, idx+1);
else if(idx >= ver2.size() && ver1[idx] != 0) return 1;
else if(idx >= ver2.size() && ver1[idx] == 0) return compareVersion(ver1, ver2, idx+1);
else if(ver1[idx] == ver2[idx]) return compareVersion(ver1, ver2, idx+1);
else if(ver1[idx] > ver2[idx]) return 1;
else return -1;
}
public:
int compareVersion(string version1, string version2) {
return compareVersion(splitVersion(version1), splitVersion(version2), 0);
}
};