Reorder List

Author: soulmachine

C++/LeetCodet题解(C++版).pdf

@@ -2598,3 +2598,75 @@ \subsubsection{相关题目}
 \begindot
 \item Linked List Cycle, 见 \S \ref{sec:Linked-List-Cycle}
 \myenddot
+
+
+\subsection{Reorder List}
+\label{sec:Reorder-List}
+
+
+\subsubsection{描述}
+Given a singly linked list $L: L_0 \rightarrow L_1 \rightarrow \cdots \rightarrow L_{n-1} \rightarrow L_n$,
+reorder it to: $L_0 \rightarrow L_n \rightarrow L_1 \rightarrow L_{n-1} \rightarrow L_2 \rightarrow L_{n-2} \rightarrow \cdots$
+
+You must do this in-place without altering the nodes' values.
+
+For example,
+Given \fn{\{1,2,3,4\}}, reorder it to \fn{\{1,4,2,3\}}.
+
+
+\subsubsection{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。
+
+
+\subsubsection{代码}
+\begin{Code}
+// LeetCode, Reorder List
+// 时间复杂度O(n)，空间复杂度O(1)
+class Solution {
+public:
+    void reorderList(ListNode *head) {
+        if (head == nullptr || head->next == nullptr) return;
+
+        ListNode *slow = head, *fast = head, *prev = nullptr;
+        while (fast && fast->next) {
+            prev = slow;
+            slow = slow->next;
+            fast = fast->next->next;
+        }
+        prev->next = nullptr; // cut at middle
+
+        slow = reverse(slow);
+
+        // merge two lists
+        ListNode *curr = head;
+        while (curr->next) {
+            ListNode *tmp = curr->next;
+            curr->next = slow;
+            slow = slow->next;
+            curr->next->next = tmp;
+            curr = tmp;
+        }
+        curr->next = slow;
+    }
+
+    ListNode* reverse(ListNode *head) {
+        if (head == nullptr || head->next == nullptr) return head;
+
+        ListNode *prev = head;
+        for (ListNode *curr = head->next, *next = curr->next; curr;
+            prev = curr, curr = next, next = next ? next->next : nullptr) {
+                curr->next = prev;
+        }
+        head->next = nullptr;
+        return prev;
+    }
+};
+\end{Code}
+
+
+\subsubsection{相关题目}
+\begindot
+\item 无
+\myenddot