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}.
Solution：后半部reverse，再依次前插
思路：
1->2->3->4->5->6
(1)快慢指针找到mid，
(2)用同206题reverse a(依次reverse)或b(前部插入reverse)方法http://www.jianshu.com/p/56353df45769 将mid后的部分reverse，变为1->2->3->6->5->4
(3)以mid分界的前半部分和后半部分：将后半部 依次插入到 前半部分中，变为1->6->2->5->3->4
屏幕快照 2017-09-12 下午1.26.23.png
Time Complexity: O(N) Space Complexity: O(1)
Solution Code：
class Solution {
public void reorderList(ListNode head) {
if(head==null || head.next==null) return;
// (1) use slow to find the middle of the list
ListNode slow = head;
ListNode fast = head;
while(fast.next != null && fast.next.next != null){
slow = slow.next;
fast = fast.next.next;
}
ListNode mid = slow;
//(2) Reverse the half after middle 1->2->3->4->5->6 to 1->2->3->6->5->4
ListNode r_head = mid.next;
ListNode cur = r_head;
ListNode post = null;
//while(cur != null & cur.next != null) {
while(cur.next != null) {
post = cur.next;
cur.next = cur.next.next;
post.next = r_head;
r_head = post;
}
mid.next = r_head; // optional
// (3) Reorder: Insert every the reversed list into first part
ListNode p1 = head;
ListNode p2 = r_head;
while(p1 != mid){
mid.next = p2.next;
p2.next = p1.next;
p1.next = p2;
p1 = p2.next;
p2 = mid.next;
}
}
}
