标签：repr hal code solution sed for contain dig col
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.
You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.
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
Example 4:
Input:version1= "1.01",version2= "1.001" Output: 0 Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”
Example 5:
Input:version1= "1.0",version2= "1.0.0" Output: 0 Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"
Solution 1:
class Solution { public int compareVersion(String version1, String version2) { if (version1 == null || version2 == null) { return 0; } String[] strArr1 = version1.split("\\."); String[] strArr2 = version2.split("\\."); int index = 0; while (index < strArr1.length && index < strArr2.length) { int cur_str1 = Integer.parseInt(strArr1[index]); int cur_str2 = Integer.parseInt(strArr2[index]); if (cur_str1 < cur_str2) { return -1; } else if (cur_str1 > cur_str2) { return 1; } index += 1; } if (index < strArr1.length) { for (int i = index; i < strArr1.length; i++) { if (Integer.parseInt(strArr1[i]) > 0) { return 1; } } } if (index < strArr2.length) { for (int j = index; j < strArr2.length; j++) { if (Integer.parseInt(strArr2[j]) > 0) { return -1; } } } return 0; } }
Solution 2:
class Solution { public int compareVersion(String version1, String version2) { String[] strArr1 = version1.split("\\."); String[] strArr2 = version2.split("\\."); int len = Math.max(strArr1.length, strArr2.length); for (int i = 0; i< len; i++) { int cur_str1 = i >= strArr1.length ? 0 : Integer.parseInt(strArr1[i]); int cur_str2 = i >= strArr2.length ? 0 : Integer.parseInt(strArr2[i]); if (cur_str1 < cur_str2) { return -1; } else if (cur_str1 > cur_str2) { return 1; } } return 0; } }
[LC] 165. Compare Version Numbers
标签：repr hal code solution sed for contain dig col
原文地址：https://www.cnblogs.com/xuanlu/p/11832642.html
