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
解题思路:
常规思路,用split()分割string, IMPORTANT : Some special characters need to be escaped while providing them a delimiters like "." and "".
然后用Integer.parseInt(x)转换成int 比较大小。注意: 1.0 和1 相等,因此判断还剩一个string 的时候,值要不等于0 才大于另一个string.
题目不难,但容易错,要小心仔细。
Java code:
public int compareVersion(String version1, String version2) { String delimiter = "\."; String[] v1 = version1.split(delimiter); String[] v2 = version2.split(delimiter); int len = Math.max(v1.length, v2.length); for(int i = 0; i< len; i++){ if(i < v1.length && i < v2.length){ if(Integer.parseInt(v1[i]) > Integer.parseInt(v2[i])) { return 1; }else if(Integer.parseInt(v1[i]) < Integer.parseInt(v2[i])){ return -1; } }else if(i < v1.length) { if(Integer.parseInt(v1[i]) > 0) { return 1; } }else if(i < v2.length){ if(Integer.parseInt(v2[i]) > 0) { return -1; } } } return 0; }
Reference:
1. http://www.programcreek.com/2014/03/leetcode-compare-version-numbers-java/