Compare two version numbers version1 and version1.
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
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
给2个版本的字符串，里面只含数字和 '.' ，比较两个版本的大小。
解法：对字符串以 '.' ，进行分拆形成数组。然后从数组的第一个元素开始比较，如果相同就比较下一个元素，如果不同v1大就返回1，v2 大就返回-1，如果比较到最后都相同就返回0。
注意：如果产生的两个数组长度不一样，例如：1.2 < 1.2.3 则要把短的后面加上'0'补齐数组长度和长的一样，然后在进行比较。
Java:
String[] arr1 = version1.split("\\.");
String[] arr2 = version2.split("\\.");
int i=0;
while(i<arr1.length || i<arr2.length){
if(i<arr1.length && i<arr2.length){
if(Integer.parseInt(arr1[i]) < Integer.parseInt(arr2[i])){
return -1;
}else if(Integer.parseInt(arr1[i]) > Integer.parseInt(arr2[i])){
return 1;
}
} else if(i<arr1.length){
if(Integer.parseInt(arr1[i]) != 0){
return 1;
}
} else if(i<arr2.length){
if(Integer.parseInt(arr2[i]) != 0){
return -1;
}
}
i++;
}
return 0;
}
Java:
public class Solution {
public int compareVersion(String version1, String version2) {
if (version1 == null || version2 == null) return 0;
String[] v1s = version1.split("\\.");
String[] v2s = version2.split("\\.");
int i = 0, j = 0, res = 0;
while (i < v1s.length || j < v2s.length) {
int ver1 = i < v1s.length ? Integer.valueOf(v1s[i++]) : 0;
int ver2 = j < v2s.length ? Integer.valueOf(v2s[j++]) : 0;
if (ver1 < ver2) return -1;
else if (ver1 > ver2) return 1;
}
return 0;
}
}
Python:
class Solution2(object):
def compareVersion(self, version1, version2):
"""
:type version1: str
:type version2: str
:rtype: int
"""
v1, v2 = version1.split("."), version2.split(".") if len(v1) > len(v2):
v2 += ['0' for _ in xrange(len(v1) - len(v2))]
elif len(v1) < len(v2):
v1 += ['0' for _ in xrange(len(v2) - len(v1))] i = 0
while i < len(v1):
if int(v1[i]) > int(v2[i]):
return 1
elif int(v1[i]) < int(v2[i]):
return -1
else:
i += 1 return 0
Python: My solution
class Solution(object):
"""
:type v1, str
:type v2, str
:rtype: int
"""
def compareVersion(self, v1, v2):
split_v1 = v1.split('.')
split_v2 = v2.split('.')
print split_v1, split_v2
n1 = len(split_v1)
n2 = len(split_v2) if n1 < n2:
for i in xrange(n2 - n1):
split_v1 += '0'
elif n1 > n2:
for i in xrange(n1 - n2):
split_v2 += '0' for i in xrange(len(split_v1)):
if split_v1[i] == split_v2[i]:
continue
if split_v1[i] > split_v2[i]:
return 1
if split_v1[i] < split_v2[i]:
return -1 return 0
类似题目：
[LeetCode] 278. First Bad Version 第一个坏版本