标签：title compareto etc span problem turn cat get dig
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
链接： http://leetcode.com/problems/compare-version-numbers/
6/16/2017
3ms, 21%
注意的问题
1. 第6，7行，string.split()对于以dot分隔的话，需要转义成\\.
2. Integer.parseInt(), Integer.toString()分别用于string->int, int->string
3. 前一部分每一块需要比较empty string
4. 如果长度不等的话，不但需要比较长度，还要比较多余长度里面是不是都是0，如果是的话2个version还是一样的。
1 public class Solution { 2 public int compareVersion(String version1, String version2) { 3 if (version1 == null || version2 == null) { 4 return 0; 5 } 6 String[] v1 = version1.split("\\."); 7 String[] v2 = version2.split("\\."); 8 9 for (int i = 0, j = 0; i < v1.length && j < v2.length; i++, j++) { 10 if (v1[i].equals("") && v2[i].equals("")) { 11 continue; 12 } 13 if (v1[i].equals("") || Integer.parseInt(v1[i]) < Integer.parseInt(v2[i])) { 14 return -1; 15 } else if (v2[i].equals("") || Integer.parseInt(v2[i]) < Integer.parseInt(v1[i])) { 16 return 1; 17 } 18 } 19 if (v1.length > v2.length) { 20 for (int i = v2.length; i < v1.length; i++) { 21 if (Integer.parseInt(v1[i]) != 0) { 22 return 1; 23 } 24 } 25 return 0; 26 } else if (v1.length < v2.length) { 27 for (int i = v1.length; i < v2.length; i++) { 28 if (Integer.parseInt(v2[i]) != 0) { 29 return -1; 30 } 31 } 32 return 0; 33 } 34 return 0; 35 } 36 }
别人的答案，2个版本都比我的好，一个是将empty string变为0，一个直接用数字来计算。
http://www.cnblogs.com/yrbbest/p/4491633.html
其他答案
https://discuss.leetcode.com/topic/6238/accepted-small-java-solution
1 public int compareVersion(String version1, String version2) { 2 String[] levels1 = version1.split("\\."); 3 String[] levels2 = version2.split("\\."); 4 5 int length = Math.max(levels1.length, levels2.length); 6 for (int i=0; i<length; i++) { 7 Integer v1 = i < levels1.length ? Integer.parseInt(levels1[i]) : 0; 8 Integer v2 = i < levels2.length ? Integer.parseInt(levels2[i]) : 0; 9 int compare = v1.compareTo(v2); 10 if (compare != 0) { 11 return compare; 12 } 13 } 14 15 return 0; 16 }
https://discuss.leetcode.com/topic/11410/my-2ms-easy-solution-with-c-c
1 int compareVersion(string version1, string version2) { 2 int i = 0; 3 int j = 0; 4 int n1 = version1.size(); 5 int n2 = version2.size(); 6 7 int num1 = 0; 8 int num2 = 0; 9 while(i<n1 || j<n2) 10 { 11 while(i<n1 && version1[i]!=‘.‘){ 12 num1 = num1*10+(version1[i]-‘0‘); 13 i++; 14 } 15 16 while(j<n2 && version2[j]!=‘.‘){ 17 num2 = num2*10+(version2[j]-‘0‘);; 18 j++; 19 } 20 21 if(num1>num2) return 1; 22 else if(num1 < num2) return -1; 23 24 num1 = 0; 25 num2 = 0; 26 i++; 27 j++; 28 } 29 30 return 0; 31 }
更多讨论
https://discuss.leetcode.com/category/173/compare-version-numbers
标签：title compareto etc span problem turn cat get dig
原文地址：http://www.cnblogs.com/panini/p/7029089.html
