• [LC] 165. Compare Version Numbers


    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"

     Solution 1:

    class Solution {
        public int compareVersion(String version1, String version2) {
            if (version1 == null || version2 == null) {
                return 0;
            }
            String[] strArr1 = version1.split("\.");
            String[] strArr2 = version2.split("\.");
            int index = 0;
            while (index < strArr1.length && index < strArr2.length) {
                int cur_str1 = Integer.parseInt(strArr1[index]);
                int cur_str2 = Integer.parseInt(strArr2[index]);
                if (cur_str1 < cur_str2) {
                    return -1;
                } else if (cur_str1 > cur_str2) {
                    return 1;
                }
                index += 1;
            }
    
            if (index < strArr1.length) {
                for (int i = index; i < strArr1.length; i++) {
                    if (Integer.parseInt(strArr1[i]) > 0) {
                        return 1;
                    }
                }
            }
            if (index < strArr2.length) {
                for (int j = index; j < strArr2.length; j++) {
                    if (Integer.parseInt(strArr2[j]) > 0) {
                        return -1;
                    }
                }
            }
            return 0;
        }
    }

    Solution 2:

    class Solution {
        public int compareVersion(String version1, String version2) {
            String[] strArr1 = version1.split("\.");
            String[] strArr2 = version2.split("\.");
            int len = Math.max(strArr1.length, strArr2.length);
            for (int i = 0; i< len; i++) {
                int cur_str1 = i >= strArr1.length ? 0 : Integer.parseInt(strArr1[i]);
                int cur_str2 = i >= strArr2.length ? 0 : Integer.parseInt(strArr2[i]);
                if (cur_str1 < cur_str2) {
                    return -1;
                } else if (cur_str1 > cur_str2) {
                    return 1;
                }
            }
            return 0;
        }
    }
  • 相关阅读:
    python完成简单购物功能
    Java对字符串进行加密
    python完成留言板功能
    webpack4 学习
    webpack学习之 vue-cli3和4
    节点操作
    new 做了什么
    从url输入网址后发生什么
    防抖节流
    BFC原理和解决方案
  • 原文地址:https://www.cnblogs.com/xuanlu/p/11832642.html
Copyright © 2020-2023  润新知