题目概述
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:
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Input: version1 = "0.1", version2 = "1.1" |
Example 2:
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Input: version1 = "1.0.1", version2 = "1" |
Example 3:
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Input: version1 = "7.5.2.4", version2 = "7.5.3" |
Example 4:
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Input: version1 = "1.01", version2 = "1.001" |
Example 5:
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Input: version1 = "1.0", version2 = "1.0.0" |
Note:
- Version strings are composed of numeric strings separated by dots
.
and this numeric strings may have leading zeroes. - Version strings do not start or end with dots, and they will not be two consecutive dots.
这题其实不难,本质上就是把字符串根据’.’进行切割后形成两个数组,然后对两个数组的元素进行逐一比较即可。
代码实现
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class : |
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