[抄题]:
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
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
以为要用stack,结果是个水题
[英文数据结构或算法,为什么不用别的数据结构或算法]:
一般就split就行了。.是特殊符号,要加双反斜杠
.split("\.")
[一句话思路]:
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
.是特殊符号,要加双反斜杠
.split("\.")
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[算法思想:迭代/递归/分治/贪心]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
[潜台词] :
class Solution { public int compareVersion(String version1, String version2) { //corner case if (version1 == null && version2 == null) return 0; //initialization: split String[] words1 = version1.split("\."); String[] words2 = version2.split("\."); for (int i = 0; i < Math.max(words1.length,words2.length); i++) { //get 2 nums int num1 = (i < words1.length) ? Integer.valueOf(words1[i]) : 0; int num2 = (i < words2.length) ? Integer.valueOf(words2[i]) : 0; //compare and return if (num1 > num2) { return 1; }else if (num1 < num2) { return -1; } } return 0; } }