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.
You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.
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
Example 4:
Input:version1= "1.01",version2= "1.001" Output: 0 Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”
Example 5:
Input:version1= "1.0",version2= "1.0.0" Output: 0 Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"
Note:
- Version strings are composed of numeric strings separated by dots
.and this numeric strings may have leading zeroes. - Version strings do not start or end with dots, and they will not be two consecutive dots.
Solution: String
Split the version string to a list of numbers, and compare two lists.
Time complexity: O(l1 + l2)
Space complexity: O(l1 + l2)
C++
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// Author: Huahua class Solution { public: int compareVersion(string version1, string version2) { const auto& v1 = parseVersion(version1); const auto& v2 = parseVersion(version2); int i1 = 0, i2 = 0; while (i1 < v1.size() || i2 < v2.size()) { int n1 = i1 < v1.size() ? v1[i1++] : 0; int n2 = i2 < v2.size() ? v2[i2++] : 0; if (n1 < n2) return -1; else if (n1 > n2) return 1; } return 0; } private: vector<int> parseVersion(const string& version) { vector<int> v; int s = 0; for (char c : version) { if (c == '.') { v.push_back(s); s = 0; } else { s = s * 10 + (c - '0'); } } v.push_back(s); return v; } }; |