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}.
基本思想:从链表中间这段,对后一段的链表作reverse,然后间隔插入到前一段的链表中。
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public void reorderList(ListNode head) {
if(head == null) return;
ListNode ln = head;
int len = 0;
while(ln!=null){
len++;
ln = ln.next;
}
int mid = (len-1)/2;
ListNode midNode = head;
while(mid!=0){
midNode = midNode.next;
mid--;
}
ListNode insert = reverse(midNode.next);
midNode.next = null;
ListNode bInsert = head;
while(insert!=null){
ListNode inext = insert.next;
ListNode bnext = bInsert.next;
bInsert.next = insert;
insert.next = bnext;
bInsert = bnext;
insert = inext;
}
}
private ListNode reverse(ListNode head){
ListNode pre = null;
ListNode cur = head;
while(cur!=null){
ListNode next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
return pre;
}
}原文地址:http://blog.csdn.net/guorudi/article/details/42242887