题目:
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
提示:
这道题难就难在考虑的是否缜密,这里列举出一些典型的测试用例:
- ["01", "1"]
- ["10.6.5", "10.6"]
- ["1.0", "1"]
代码:
我自己的实现方法(比较臃肿,不推荐):
class Solution { public: int compareVersion(string version1, string version2) { int v1, v2, pos1, pos2; while (true) { if (version1.size() == 0) { if (version2.size() == 0) return 0; else version1 = "0"; } else { if (version2.size() == 0) version2 = "0"; } pos1 = version1.find("."); if (pos1 > -1) { v1 = atoi(version1.substr(0, pos1).c_str()); version1 = version1.substr(pos1 + 1, version1.size() - 1); } else { v1 = atoi(version1.c_str()); version1 = ""; } pos2 = version2.find("."); if (pos2 > -1) { v2 = atoi(version2.substr(0, pos2).c_str()); version2 = version2.substr(pos2 + 1, version2.size() - 1); } else { v2 = atoi(version2.c_str()); version2 = ""; } if (v1 > v2) return 1; if (v1 < v2) return -1; } } };
论坛里某大神的方法(非常简洁,推荐):
class Solution { public: int compareVersion(string version1, string version2) { for (auto& w : version1) if (w == '.') w = ' '; for (auto& w : version2) if (w == '.') w = ' '; istringstream s1(version1), s2(version2); while (true) { int n1, n2; if (!(s1 >> n1)) n1 = 0; if (!(s2 >> n2)) n2 = 0; if (!s1 && !s2) return 0; if (n1 < n2) return -1; if (n1 > n2) return 1; } } };