62. 链表重排[Reorder List]
【本文链接】
http://www.cnblogs.com/hellogiser/p/reorder-list.html
【题目】
Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You must do this in-place without altering the nodes' values.
For example,
Given {1,2,3,4,5,6,7}, reorder it to {1,7,2,6,3,5,4}.
【分析】
题目思路比较直接：
(1)找到链表的中间节点，把链表划分成2个子链表；如果原链表长度为奇数，那么第一个子链表的长度多1；
(2)翻转第二个子链表；
(3)交叉合并两个子链表。
例如{1,2,3,4,5,6,7}
(1)找到链表的中间节点为4，把链表划分成2个子链表：{1,2,3,4}和{5,6,7}；
(2)翻转第二个子链表得到{7,6,5}
(3)交叉合并{1,2,3,4}和{7,6,5}得到{1,7,2,6,3,5,4}
【代码】
C++ Code
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// 62_ReorderList.cpp : Defines the entry point for the console application.
// /* version: 1.0 author: hellogiser blog: http://www.cnblogs.com/hellogiser date: 2014/5/30 */ #include "stdafx.h" struct ListNode { int value; ListNode *next; }; // find middle node of list ListNode *FindMiddleNode(ListNode *head) { if(NULL == head) return NULL; ListNode *fast = head, *slow = head; while(fast != NULL && fast->next != NULL) { // move fast 2 steps fast = fast->next->next; if (fast == NULL) break; // move slow 1 step slow = slow->next; } return slow; } // reverse list ListNode *ReverseList(ListNode *head) { if(NULL == head || NULL == head->next) return head; ListNode *prev = NULL, *cur = head, *next = NULL; while(cur != NULL) { // save next next = cur->next; // reverse cur->next = prev; // update prev and cur prev = cur; cur = next; } return prev; } // cross merge list ListNode *CrossMergeList(ListNode *head1, ListNode *head2) { if(NULL == head1) return head2; else if (NULL == head2) return head1; ListNode *node1 = head1, *node2 = head2; while(node2 != NULL) { ListNode *temp1 = node1->next; ListNode *temp2 = node2->next; node1->next = node2; node2->next = temp1; // update node1 node2 node1 = temp1; node2 = temp2; } return head1; } // reorder list ListNode *ReOrderList(ListNode *head) { if(NULL == head || NULL == head->next) return head; // find middle node of list ListNode *middle = FindMiddleNode(head); // split into 2 lists ListNode *head1 = head; ListNode *head2 = middle->next; // detach the 2 lists middle->next = NULL; // reverse list2 head2 = ReverseList(head2); // cross merge 2 lists return CrossMergeList(head1, head2); } |
【参考】
http://blog.csdn.net/whuwangyi/article/details/14146461
【本文链接】
