{"id":5309,"date":"2019-07-20T21:42:41","date_gmt":"2019-07-21T04:42:41","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=5309"},"modified":"2019-07-21T09:28:07","modified_gmt":"2019-07-21T16:28:07","slug":"leetcode-1128-number-of-equivalent-domino-pairs","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/hashtable\/leetcode-1128-number-of-equivalent-domino-pairs\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1128. Number of Equivalent Domino Pairs"},"content":{"rendered":"\n

Given a list of dominoes<\/code>, dominoes[i] = [a, b]<\/code> is equivalent<\/em> to dominoes[j] = [c, d]<\/code> if and only if either (a==c<\/code> and b==d<\/code>), or (a==d<\/code> and b==c<\/code>) – that is, one domino can be rotated to be equal to another domino.<\/p>\n\n\n\n

Return the number of pairs (i, j)<\/code> for which 0 <= i < j < dominoes.length<\/code>, and dominoes[i]<\/code> is equivalent to dominoes[j]<\/code>.<\/p>\n\n\n\n

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

Input:<\/strong> dominoes = [[1,2],[2,1],[3,4],[5,6]]\nOutput:<\/strong> 1\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= dominoes.length <= 40000<\/code><\/li>
  • 1 <= dominoes[i][j] <= 9<\/code><\/li><\/ul>\n\n\n\n
    Solution: HashTable<\/strong><\/h2>\n\n\n\n
    Count how many times each key occurred so far.<\/p>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(100)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \nclass Solution {\npublic:\n  int numEquivDominoPairs(vector>& dominoes) {\n    vector m(100);\n    int ans = 0;\n    for (const auto& d : dominoes) {\n      int k1 = d[0] * 10 + d[1];\n      int k2 = d[1] * 10 + d[0];\n      ans += m[k1];\n      if (k1 != k2) ans += m[k2];\n      ++m[k1];\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    Given a list of dominoes, dominoes[i] = [a, b] is equivalent to dominoes[j] = [c, d] if and only if either (a==c and b==d), or (a==d and b==c) – that is, one domino can be rotated…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[70],"tags":[222,82],"class_list":["post-5309","post","type-post","status-publish","format-standard","hentry","category-hashtable","tag-easy","tag-hashtable","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5309","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=5309"}],"version-history":[{"count":4,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5309\/revisions"}],"predecessor-version":[{"id":5324,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5309\/revisions\/5324"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=5309"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=5309"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=5309"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}