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[Leetcode]165. Compare Version Numbers(C++)

题目描述

题目链接:165. Compare Version Numbers

Given two version numbers, version1 and version2, 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 example 2.5.33 and 0.1 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 and 001 are considered equal. If a version number does not specify a revision at an index, then treat the revision as 0. For example, version 1.0 is less than version 1.1 because their revision 0s are the same, but their revision 1s are 0 and 1 respectively, and 0 < 1.

Return the following:

  • If version1 < version2, return -1.
  • If version1 > version2, return 1.
  • 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 and version2 only contain digits and '.'.
  • version1 and version2 are valid version numbers.
  • All the given revisions in version1 and version2 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:

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#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