Reorder List
Reorder List
Total Accepted: 15732 Total Submissions: 78041 My SubmissionsGiven 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}.
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这道题目，第一次见会感觉比较麻烦，以后见到了，就知道套路。先把链表截断，接着把链表的下半段反转，然后再把两个链表合并。截断链表的位置需要根据链表的长度来判断，奇数偶数稍有不同，自己找规律。整个过程没有改变链表节点的值，只是改变了其顺序，并且空间也是常数。
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
/**
* @param args
*/
/*
public static void main(String[] args) {
ListNode res,node = new ListNode(1);
res = node;
for(int i=2;i<=4;i++)
{
ListNode n = new ListNode(i);
node.next = n;
node = node.next;
}
Solution ss = new Solution();
ss.reorderList(res);
System.out.println(res);
}
*/
public void reorderList(ListNode head){
int len = getLen(head);
if(len<3)
{
return;
}
int cut = (len+1)/2,count=1;
ListNode h1 = head,h2 = null,h3 = null,h4=null;
while(count<cut)
{
h1 = h1.next;
count++;
}
h2 = h1.next;
h1.next = null;
h2 = reverseList(h2);
ListNode res = new ListNode(-1);
h1 = head;
while(h1!=null&&h2!=null)
{
h3 = h1.next;
h1.next = h2;
h4 = h2.next;
h2.next=h3;
h1 = h3;
h2 = h4;
}
return ;
}
ListNode reverseList(ListNode head)
{
ListNode n1,n2,n3;
if(head==null||head.next==null)
{
return head;
}
n1 = head;
n2 = n1.next;
n1.next = null;
while(n2!=null)
{
n3 = n2.next;
n2.next = n1;
n1 = n2;
n2 = n3;
}
return n1;
}
int getLen(ListNode head)
{
int count = 0;
while(head!=null){
count++;
head = head.next;
}
return count;
}
}