LeetCode算法入门- Compare Version Numbers -day14

1. 题目描述：
   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:

Input: version1 = “0.1”, version2 = “1.1”
Output: -1

Example 2:

Input: version1 = “1.0.1”, version2 = “1”
Output: 1

Example 3:

Input: version1 = “7.5.2.4”, version2 = “7.5.3”
Output: -1

Example 4:

Input: version1 = “1.01”, version2 = “1.001”
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”

Example 5:

Input: version1 = “1.0”, version2 = “1.0.0”
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to “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.

2. 题意分析：

该题目的意思是比较【版本号】。版本号只是有“数字”和”.“组成

if version1 > version2 return 1
if version1 < version2 return -1
其他情况返回0.

3. Java实现：
   方法一：通过split函数分割开，然后将其转换为int类型，然后一一进行比较

class Solution {public int compareVersion(String version1, String version2) {//进行分割，.*|分割的时候要加上\\String[] str1 = version1.split("\\.");String[] str2 = version2.split("\\.");//取最长的字符串作为基准int maxLen = Math.max(str1.length , str2.length);for(int i = 0; i < maxLen; i++){int temp1 = 0; int temp2 = 0;if(i < str1.length){temp1 = Integer.parseInt(str1[i]);}if(i < str2.length){temp2 = Integer.parseInt(str2[i]);}if(temp1 < temp2)return -1;else if(temp1 > temp2)return 1;}return 0;}
}

方法二，也可以不使用split函数，直接循环判断字符串的方法，如果为“.”，则向右移动，单个单个进行比较

class Solution {public int compareVersion(String version1, String version2) {int i = 0;int j = 0;while(i < version1.length() || j <version2.length()){int temp1 = 0;int temp2 = 0;//如果指向.的话，i++，向右移动while(i < version1.length() && version1.charAt(i) != '.'){//以防版本号为两位数11.1.2temp1 = temp1 * 10 + (version1.charAt(i) - '0');i++;}i++;while(j < version2.length() && version2.charAt(j) != '.'){//通过(version2.charAt(j) - '0')来获取其int值temp2 = temp2 * 10 + (version2.charAt(j) - '0');j++;}j++;if(temp1 > temp2)return 1;else if(temp1 < temp2){return -1;}}return 0;}
}