{"id":2477,"date":"2018-04-12T22:23:44","date_gmt":"2018-04-13T05:23:44","guid":{"rendered":"http:\/\/zxi.mytechroad.com\/blog\/?p=2477"},"modified":"2018-04-12T22:23:57","modified_gmt":"2018-04-13T05:23:57","slug":"leetcode-328-odd-even-linked-list","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/list\/leetcode-328-odd-even-linked-list\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 328. Odd Even Linked List"},"content":{"rendered":"
\n
\n

Problem<\/strong><\/h1>\n

\u9898\u76ee\u5927\u610f\uff1a\u7ed9\u4f60\u4e00\u4e2a\u94fe\u8868\uff0c\u628a\u6240\u6709\u5947\u6570\u4f4d\u7f6e\u7684\u8282\u70b9\u4e32\u5728\u4e00\u8d77\uff0c\u540e\u9762\u8ddf\u7740\u6240\u6709\u4e32\u5728\u4e00\u8d77\u7684\u5076\u6570\u4f4d\u7f6e\u7684\u8282\u70b9\u3002<\/p>\n

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.<\/p>\n

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.<\/p>\n

Example:<\/b>
\nGiven\u00a01->2->3->4->5->NULL<\/code>,
\nreturn\u00a01->3->5->2->4->NULL<\/code>.<\/p>\n

Note:<\/b>
\nThe relative order inside both the even and odd groups should remain as it was in the input.
\nThe first node is considered odd, the second node even and so on …<\/p>\n

Credits:<\/b>
\nSpecial thanks to\u00a0@DjangoUnchained<\/a>\u00a0for adding this problem and creating all test cases.<\/p>\n<\/div>\n<\/div>\n

Solution<\/h1>\n

Time complexity: O(n)<\/p>\n

Space complexity: O(1)<\/p>\n

C++<\/p>\n

\/\/ Author: Huahua\r\n\/\/ Running time: 21 ms\r\nclass Solution {\r\npublic:\r\n  ListNode* oddEvenList(ListNode* head) {\r\n    if (head == nullptr) return head;\r\n    ListNode dummy_odd(0);\r\n    ListNode dummy_even(0);\r\n    ListNode* prev_odd = &dummy_odd;\r\n    ListNode* prev_even = &dummy_even;\r\n    int index = 0;\r\n    while (head) {      \r\n      auto next = head->next;\r\n      head->next = nullptr; \/\/ important\r\n      if (index++ & 1) {\r\n        prev_even->next = head;\r\n        prev_even = head;\r\n      } else {\r\n        prev_odd->next = head;\r\n        prev_odd = head;\r\n      }\r\n      head = next;\r\n    }\r\n    prev_odd->next = dummy_even.next;\r\n    return dummy_odd.next;\r\n  }\r\n};<\/pre>\n
V2<\/p>\n
\/\/ Author: Huahua\r\n\/\/ Running time: 18 ms\r\nclass Solution {\r\npublic:\r\n  ListNode* oddEvenList(ListNode* head) {\r\n    if (head == nullptr) return head;\r\n    vector<ListNode> heads(2, ListNode(0));    \r\n    vector<ListNode*> prevs{&heads[0], &heads[1]};\r\n    int index = 0;\r\n    while (head) {\r\n      auto next = head->next;\r\n      head->next = nullptr; \/\/ important      \r\n      prevs[index]->next = head;\r\n      prevs[index] = head;\r\n      head = next;\r\n      index ^= 1;\r\n    }\r\n    prevs[0]->next = heads[1].next;\r\n    return heads[0].next;\r\n  }\r\n};<\/pre>\n
Java<\/p>\n
\/\/ Author: Huahua\r\n\/\/ Running time: 1 ms\r\nclass Solution {\r\n  public ListNode oddEvenList(ListNode head) {\r\n    ListNode[] heads = new ListNode[]{new ListNode(0), new ListNode(0)};\r\n    ListNode[] prevs = new ListNode[]{heads[0], heads[1]};\r\n    int index = 0;\r\n    while (head != null) {\r\n      ListNode next = head.next;\r\n      head.next = null;\r\n      prevs[index].next = head;\r\n      prevs[index] = prevs[index].next;\r\n      head = next;\r\n      index ^= 1;\r\n    }\r\n    prevs[0].next = heads[1].next;\r\n    return heads[0].next;\r\n  }\r\n}<\/pre>\n
 <\/p>\n
Python3<\/p>\n
\"\"\"\r\nAuthor: Huahua\r\nRunning time: 56 ms\r\n\"\"\"\r\nclass Solution:\r\n  def oddEvenList(self, head):\r\n    heads = [ListNode(0), ListNode(0)]\r\n    prevs = [heads[0], heads[1]]\r\n    index = 0\r\n    while head:\r\n      nxt = head.next\r\n      head.next = None\r\n      prevs[index].next = head\r\n      prevs[index] = prevs[index].next\r\n      head = nxt\r\n      index ^= 1\r\n    prevs[0].next = heads[1].next\r\n    return heads[0].next<\/pre>\n
 <\/p>\n","protected":false},"excerpt":{"rendered":"
Problem \u9898\u76ee\u5927\u610f\uff1a\u7ed9\u4f60\u4e00\u4e2a\u94fe\u8868\uff0c\u628a\u6240\u6709\u5947\u6570\u4f4d\u7f6e\u7684\u8282\u70b9\u4e32\u5728\u4e00\u8d77\uff0c\u540e\u9762\u8ddf\u7740\u6240\u6709\u4e32\u5728\u4e00\u8d77\u7684\u5076\u6570\u4f4d\u7f6e\u7684\u8282\u70b9\u3002 Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the…<\/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":[295,83,177],"class_list":["post-2477","post","type-post","status-publish","format-standard","hentry","category-list","tag-dummy-head","tag-list","tag-medium","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/2477","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=2477"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/2477\/revisions"}],"predecessor-version":[{"id":2479,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/2477\/revisions\/2479"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=2477"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=2477"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=2477"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}