Reorder List

描述

Given a singly linked list L:L0→L1→⋯→Ln−1→LnL: L0 \rightarrow L_1 \rightarrow \cdots \rightarrow L{n-1} \rightarrow Ln,reorder it to: L0→Ln→L1→Ln−1→L2→Ln−2→⋯L_0 \rightarrow L_n \rightarrow L_1 \rightarrow L{n-1} \rightarrow L2 \rightarrow L{n-2} \rightarrow \cdots

You must do this in-place without altering the nodes' values.

For example,Given {1,2,3,4}, reorder it to {1,4,2,3}.

分析

题目规定要in-place，也就是说只能使用O(1)的空间。

可以找到中间节点，断开，把后半截单链表reverse一下，再合并两个单链表。

代码

  1. // Reorder List
  2. // 时间复杂度O(n)，空间复杂度O(1)
  3. class Solution {
  4. public:
  5.     void reorderList(ListNode *head) {
  6.         if (head == nullptr || head->next == nullptr) return;
  8.         ListNode *slow = head, *fast = head, *prev = nullptr;
  9.         while (fast && fast->next) {
  10.             prev = slow;
  11.             slow = slow->next;
  12.             fast = fast->next->next;
  13.         }
  14.         prev->next = nullptr; // cut at middle
  16.         slow = reverse(slow);
  18.         // merge two lists
  19.         ListNode *curr = head;
  20.         while (curr->next) {
  21.             ListNode *tmp = curr->next;
  22.             curr->next = slow;
  23.             slow = slow->next;
  24.             curr->next->next = tmp;
  25.             curr = tmp;
  26.         }
  27.         curr->next = slow;
  28.     }
  30.     ListNode* reverse(ListNode *head) {
  31.         if (head == nullptr || head->next == nullptr) return head;
  33.         ListNode *prev = head;
  34.         for (ListNode *curr = head->next, *next = curr->next; curr;
  35.             prev = curr, curr = next, next = next ? next->next : nullptr) {
  36.                 curr->next = prev;
  37.         }
  38.         head->next = nullptr;
  39.         return prev;
  40.     }
  41. };

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/cpp/linear-list/linked-list/reorder-list.html