Problem
Given a singly linked list L: L0→L1→…→Ln-1→Ln, reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You may not modify the values in the list's nodes, only nodes itself may be changed.
Example 1:
Given 1->2->3->4, reorder it to 1->4->2->3.
Example 2:
Given 1->2->3->4->5, reorder it to 1->5->2->4->3.
Solution
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {void} Do not return anything, modify head in-place instead.
*/
var reorderList = function(head) {
if (!head || !head.next || !head.next.next) return;
// find mid
var mid = null;
var fast = head;
var slow = head;
while (fast.next && fast.next.next && slow.next) {
slow = slow.next;
fast = fast.next.next;
}
mid = slow;
// reverse the later part
var now = mid.next.next;
var second = mid.next;
var tmp = null;
second.next = null;
while (now) {
tmp = now.next;
now.next = second;
second = now;
now = tmp;
}
mid.next = second;
// insert one after another
var before = head;
var after = mid.next;
mid.next = null;
while (after) {
tmp = before.next;
before.next = after;
before = tmp;
tmp = after.next;
after.next = before;
after = tmp
}
};
Explain:
比如 1->2->3->4->5->6
,分三步
- 找到
mid = 3
- 翻转后半部分,变成
1->2->3->6->5->4
- 后半部分依次插入到前半部分
Complexity:
- Time complexity : O(n).
- Space complexity : O(1).