{"id":3351,"date":"2018-07-28T21:35:52","date_gmt":"2018-07-29T04:35:52","guid":{"rendered":"http:\/\/zxi.mytechroad.com\/blog\/?p=3351"},"modified":"2018-07-28T21:36:05","modified_gmt":"2018-07-29T04:36:05","slug":"leetcode-876-middle-of-the-linked-list","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/list\/leetcode-876-middle-of-the-linked-list\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 876. Middle of the Linked List"},"content":{"rendered":"

Problem<\/h1>\n

Given a non-empty, singly\u00a0linked list with head node\u00a0head<\/code>, return\u00a0a\u00a0middle node of linked list.<\/p>\n

If there are two middle nodes, return the second middle node.<\/p>\n

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

Input: <\/strong>[1,2,3,4,5]<\/span>\r\nOutput: <\/strong>Node 3 from this list (Serialization: [3,4,5]<\/span>)\r\nThe returned node has value 3.  (The judge's serialization of this node is [3,4,5]).\r\nNote that we returned a ListNode object ans, such that:\r\nans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.\r\n<\/pre>\n
\n
Example 2:<\/strong><\/p>\n
Input: <\/strong>[1,2,3,4,5,6]<\/span>\r\nOutput: <\/strong>Node 4 from this list (Serialization: [4,5,6]<\/span>)\r\nSince the list has two middle nodes with values 3 and 4, we return the second one.\r\n<\/pre>\n
 <\/p>\n
Note:<\/strong><\/p>\n
    \n
  • The number of nodes in the given list will be between\u00a01<\/code>\u00a0and\u00a0100<\/code>.<\/li>\n<\/ul>\n
    Solution: Slow + Fast Pointers<\/strong><\/h1>\n
    Time complexity: O(n)<\/p>\n
    Space complexity: O(1)<\/p>\n
    \/\/ Author: Huahua\r\n\/\/ Running time: 0 ms\r\nclass Solution {\r\npublic:\r\n  ListNode* middleNode(ListNode* head) {\r\n    if (!head || !head->next) return head;\r\n    auto slow = head;\r\n    auto fast = head;\r\n    while (fast && fast->next) {\r\n      slow = slow->next;\r\n      fast = fast->next->next;\r\n    }\r\n    return slow;\r\n  }\r\n};<\/pre>\n
     <\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"
    Problem Given a non-empty, singly\u00a0linked list with head node\u00a0head, return\u00a0a\u00a0middle node of linked list. If there are two middle nodes, return the second middle node.…<\/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":[222,83,358],"class_list":["post-3351","post","type-post","status-publish","format-standard","hentry","category-list","tag-easy","tag-list","tag-middle","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/3351","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=3351"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/3351\/revisions"}],"predecessor-version":[{"id":3353,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/3351\/revisions\/3353"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=3351"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=3351"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=3351"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}