题目描述
题目链接:165. Compare Version Numbers
Given two version numbers,
version1
andversion2
, compare them.Version numbers consist of one or more revisions joined by a dot
'.'
. Each revision consists of digits and may contain leading zeros. Every revision contains at least one character. Revisions are 0-indexed from left to right, with the leftmost revision being revision 0, the next revision being revision 1, and so on. For example2.5.33
and0.1
are valid version numbers.To compare version numbers, compare their revisions in left-to-right order. Revisions are compared using their integer value ignoring any leading zeros. This means that revisions
1
and001
are considered equal. If a version number does not specify a revision at an index, then treat the revision as0
. For example, version1.0
is less than version1.1
because their revision 0s are the same, but their revision 1s are0
and1
respectively, and0
<1
.Return the following:
- If
version1
<version2
, return-1
.- If
version1
>version2
, return1
.- Otherwise, return
0
.
例子
例子 1
Input: version1 = “1.01”, version2 = “1.001” Output: 0 Explanation: Ignoring leading zeroes, both “01” and “001” represent the same integer “1”.
例子 2
Input: version1 = “1.0”, version2 = “1.0.0” Output: 0 Explanation: version1 does not specify revision 2, which means it is treated as “0”.
例子 3
Input: version1 = “0.1”, version2 = “1.1” Output: -1 Explanation: version1’s revision 0 is “0”, while version2’s revision 0 is “1”. 0 < 1, so version1 < version2.
例子 3
Input: version1 = “1.0.1”, version2 = “1” Output: 1
例子 3
Input: version1 = “7.5.2.4”, version2 = “7.5.3” Output: -1
Constraints
1 <= version1.length, version2.length <= 500
version1
andversion2
only contain digits and'.'
.version1
andversion2
are valid version numbers.- All the given revisions in
version1
andversion2
can be stored in a 32-bit integer.
解题思路
这道题主要考察对字符串的基本操作。通过两个指针分别遍历两个版本号,以 .
为分隔点,取出每个小版本号进行比较即可。注意这里结束遍历的条件必须是两个字符串都遍历完。中途只要遇到一个版本号不同就可以根据比较情况返回结果,代码如下:
#include <string>
class Solution {
public:
int compareVersion(std::string version1, std::string version2) {
int rev1 = 0, rev2 = 0;
int ptr1 = 0, ptr2 = 0;
while (ptr1 < version1.length() || ptr2 < version2.length()) {
rev1 = 0;
while (ptr1 < version1.length() && version1[ptr1] != '.') {
rev1 *= 10;
rev1 += version1[ptr1] - '0';
ptr1++;
}
rev2 = 0;
while (ptr2 < version2.length() && version2[ptr2] != '.') {
rev2 *= 10;
rev2 += version2[ptr2] - '0';
ptr2++;
}
if (rev1 != rev2) {
return rev1 < rev2 ? -1 : 1;
}
ptr1++;
ptr2++;
}
return 0;
}
};
- 时间复杂度: O(n)
- 空间复杂度: O(1)
GitHub 代码同步地址: 165.CompareVersionNumbers.cpp
其他题目: GitHub: Leetcode-C++-Solution 博客: Leetcode-Solutions