{"id":6435,"date":"2020-03-09T01:06:39","date_gmt":"2020-03-09T08:06:39","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=6435"},"modified":"2020-03-09T01:06:59","modified_gmt":"2020-03-09T08:06:59","slug":"leetcode-165-compare-version-numbers","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/string\/leetcode-165-compare-version-numbers\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 165. Compare Version Numbers"},"content":{"rendered":"\n

Compare two version numbers version1<\/em> and version2<\/em>.
If version1<\/em> > version2<\/em><\/code> return 1;<\/code> if version1<\/em> < version2<\/em><\/code> return -1;<\/code>otherwise return 0<\/code>.<\/p>\n\n\n\n

You may assume that the version strings are non-empty and contain only digits and the .<\/code> character.<\/p>\n\n\n\n

The .<\/code> character does not represent a decimal point and is used to separate number sequences.<\/p>\n\n\n\n

For instance, 2.5<\/code> 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.<\/p>\n\n\n\n

You may assume the default revision number for each level of a version number to be 0<\/code>. For example, version number 3.4<\/code> has a revision number of 3<\/code> and 4<\/code> for its first and second level revision number. Its third and fourth level revision number are both 0<\/code>.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> version1<\/em><\/code> = \"0.1\", version2<\/em><\/code> = \"1.1\"\nOutput:<\/strong> -1<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input: <\/strong>version1<\/em><\/code> = \"1.0.1\", version2<\/em><\/code> = \"1\"\nOutput:<\/strong> 1<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> version1<\/em><\/code> = \"7.5.2.4\", version2<\/em><\/code> = \"7.5.3\"\nOutput:<\/strong> -1<\/pre>\n\n\n\n
Example 4:<\/strong><\/p>\n\n\n\n
Input:<\/strong> version1<\/em><\/code> = \"1.01\", version2<\/em><\/code> = \"1.001\"\nOutput:<\/strong> 0\nExplanation:<\/strong> Ignoring leading zeroes, both \u201c01\u201d and \u201c001\" represent the same number \u201c1\u201d<\/pre>\n\n\n\n
Example 5:<\/strong><\/p>\n\n\n\n
Input:<\/strong> version1<\/em><\/code> = \"1.0\", version2<\/em><\/code> = \"1.0.0\"\nOutput:<\/strong> 0\nExplanation:<\/strong> The first version number does not have a third level revision number, which means its third level revision number is default to \"0\"<\/pre>\n\n\n\n
Note:<\/strong><\/p>\n\n\n\n
  1. Version strings are composed of numeric strings separated by dots .<\/code> and this numeric strings may<\/strong> have leading zeroes.<\/li>
  2. Version strings do not start or end with dots, and they will not be two consecutive dots.<\/li><\/ol>\n\n\n\n
    Solution: String<\/p>\n\n\n\n
    Split the version string to a list of numbers, and compare two lists.<\/p>\n\n\n\n
    Time complexity: O(l1 + l2)
    Space complexity: O(l1 + l2)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int compareVersion(string version1, string version2) {\n    const auto& v1 = parseVersion(version1);\n    const auto& v2 = parseVersion(version2);\n    int i1 = 0, i2 = 0;\n    while (i1 < v1.size() || i2 < v2.size()) {    \n      int n1 = i1 < v1.size() ? v1[i1++] : 0;\n      int n2 = i2 < v2.size() ? v2[i2++] : 0;\n      if (n1 < n2) return -1;\n      else if (n1 > n2) return 1;\n    }\n    return 0;\n  }\nprivate:\n  vector parseVersion(const string& version) {\n    vector v;\n    int s = 0;\n    for (char c : version) {\n      if (c == '.') {\n        v.push_back(s);\n        s = 0;\n      } else {\n        s = s * 10 + (c - '0');\n      }\n    }\n    v.push_back(s);\n    return v;\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
    <\/p>\n","protected":false},"excerpt":{"rendered":"
    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…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[47],"tags":[93,567,177,4],"class_list":["post-6435","post","type-post","status-publish","format-standard","hentry","category-string","tag-conversion","tag-dot","tag-medium","tag-string","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/6435","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=6435"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/6435\/revisions"}],"predecessor-version":[{"id":6437,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/6435\/revisions\/6437"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=6435"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=6435"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=6435"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}