标签：des style blog http color io
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}, reorder it to{1,4,2,3}.
算法思路：
设置快慢指针将前半段与后半段分开，然后将后半段逆序，再逐个插入前半段，时间复杂度O(n)，空间复杂度不定
思路1：
后半段的逆序，设置三指针，在原list基础上逆序，空间复杂度O(1)
后面还有一些题会用这个思路，这里就不实现了。
思路2：
后半段的逆序，借助栈，空间复杂度O(n)，代码简单
代码如下：
1 public class Solution { 2 public void reorderList(ListNode head) { 3 if(head == null || head.next == null) return ; 4 ListNode hhead = new ListNode(0); 5 hhead.next = head; 6 ListNode fast = hhead; 7 ListNode slow = hhead; 8 while(fast != null && fast.next != null){ 9 fast = fast.next.next; 10 slow = slow.next; 11 } 12 ListNode stackPart = slow.next; 13 slow.next = null; 14 Stack<ListNode> stack = new Stack<ListNode>(); 15 while(stackPart != null){ 16 stack.push(stackPart); 17 stackPart = stackPart.next; 18 } 19 ListNode insert = head; 20 while(!stack.isEmpty()){ 21 ListNode tem = stack.pop(); 22 tem.next = insert.next; 23 insert.next = tem; 24 insert = tem.next; 25 } 26 } 27 }
[leetcode]Reorder List,布布扣,bubuko.com
标签：des style blog http color io
原文地址：http://www.cnblogs.com/huntfor/p/3858468.html
