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  • 165. Compare Version Numbers比较版本号的大小

    [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;
        }
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/immiao0319/p/9402277.html
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