{"id":9522,"date":"2022-03-02T06:51:51","date_gmt":"2022-03-02T14:51:51","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=9522"},"modified":"2022-03-02T06:52:52","modified_gmt":"2022-03-02T14:52:52","slug":"leetcode-2181-merge-nodes-in-between-zeros","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/list\/leetcode-2181-merge-nodes-in-between-zeros\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 2181. Merge Nodes in Between Zeros"},"content":{"rendered":"\n

You are given the head<\/code> of a linked list, which contains a series of integers separated<\/strong> by 0<\/code>‘s. The beginning<\/strong> and end<\/strong> of the linked list will have Node.val == 0<\/code>.<\/p>\n\n\n\n

For every <\/strong>two consecutive 0<\/code>‘s, merge<\/strong> all the nodes lying in between them into a single node whose value is the sum<\/strong> of all the merged nodes. The modified list should not contain any 0<\/code>‘s.<\/p>\n\n\n\n

Return the<\/em> head<\/code> of the modified linked list<\/em>.<\/p>\n\n\n\n

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

\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> head = [0,3,1,0,4,5,2,0]\nOutput:<\/strong> [4,11]\nExplanation:<\/strong> \nThe above figure represents the given linked list. The modified list contains\n- The sum of the nodes marked in green: 3 + 1 = 4.\n- The sum of the nodes marked in red: 4 + 5 + 2 = 11.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> head = [0,1,0,3,0,2,2,0]\nOutput:<\/strong> [1,3,4]\nExplanation:<\/strong> \nThe above figure represents the given linked list. The modified list contains\n- The sum of the nodes marked in green: 1 = 1.\n- The sum of the nodes marked in red: 3 = 3.\n- The sum of the nodes marked in yellow: 2 + 2 = 4.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • The number of nodes in the list is in the range [3, 2 * 105<\/sup>]<\/code>.<\/li>
  • 0 <= Node.val <= 1000<\/code><\/li>
  • There are no<\/strong> two consecutive nodes with Node.val == 0<\/code>.<\/li>
  • The beginning<\/strong> and end<\/strong> of the linked list have Node.val == 0<\/code>.<\/li><\/ul>\n\n\n\n
    Solution: List<\/strong><\/h2>\n\n\n\n
    Skip the first zero, replace every zero node with the sum of values of its previous nodes.<\/p>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  ListNode* mergeNodes(ListNode* head) {\n    ListNode dummy;    \n    head = head->next;\n    for (ListNode* prev = &dummy; head; head = head->next) {\n      int sum = 0;\n      while (head->val != 0) {\n        sum += head->val;\n        head = head->next;\n      }\n      prev->next = head;\n      head->val = sum;\n      prev = head;      \n    }\n    return dummy.next;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    You are given the head of a linked list, which contains a series of integers separated by 0‘s. The beginning and end of the linked list will have Node.val == 0. For every two consecutive 0‘s, merge all the nodes…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[50],"tags":[83,177,129,767],"class_list":["post-9522","post","type-post","status-publish","format-standard","hentry","category-list","tag-list","tag-medium","tag-merge","tag-skip","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9522","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=9522"}],"version-history":[{"count":3,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9522\/revisions"}],"predecessor-version":[{"id":9525,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9522\/revisions\/9525"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=9522"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=9522"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=9522"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}