Questions:

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

Example 3:

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

Thinking:

  • Method 1:
    • 要注意在使用split的时候”.”需要使用转义字符”\.” 
      class Solution {
  public int compareVersion(String version1, String version2) {
  String[] arr1 = version1.split("\\.");
  String[] arr2 = version2.split("\\.");
  int p1 = 0, p2 = 0;
  while(p1 < arr1.length || p2 < arr2.length){
      if(p1 >= arr1.length && p2 < arr2.length){
          if(Integer.valueOf(arr2[p2]) != 0)
              return -1;
          else{
              p2++;continue;
          }
      }
      if(p2 >= arr2.length && p1 < arr1.length){
          if(Integer.valueOf(arr1[p1]) != 0)
              return 1;
          else{ p1++; continue;}
      }
      if(p1 < arr1.length && p2 < arr2.length){
          if(Integer.valueOf(arr1[p1]) == Integer.valueOf(arr2[p2])){
              p1++;
              p2++;
          }else if(Integer.valueOf(arr1[p1]) > Integer.valueOf(arr2[p2]))
              return 1;
          else if(Integer.valueOf(arr1[p1]) < Integer.valueOf(arr2[p2]))
              return -1;
          }
  }
  return 0;
  }
}

二刷

  1. 还是要注意使用转义字符，因为.在正则中是单字符匹配。
  2. 从字符串转成整形有两种方法，一种是Integer.valueOf()，第二种是Integer.parseInt(),似乎valueOf方法更快一些。 
    class Solution {
 public int compareVersion(String version1, String version2) {
     String[] arr1 = version1.split("\\.");        
     String[] arr2 = version2.split("\\.");
     int len1 = arr1.length, len2 = arr2.length;
     int index1 = 0, index2 = 0;
     while(index1 < len1 && index2 < len2){
         Integer token1 = Integer.valueOf(arr1[index1++]);
         Integer token2 = Integer.valueOf(arr2[index2++]);
         if(token1 < token2) return -1;
         else if(token1 > token2) return 1;
     }
     if(index1 == len1){
         while(index2 < len2){
             if(Integer.valueOf(arr2[index2++]) != 0)
                 return -1;
         }
     }else{
         while(index1 < len1){
             if(Integer.valueOf(arr1[index1++]) != 0)
                 return 1;
         }
     }
     return 0;
 }
}