JS Leecode 143.Reorder List

143.Reorder List

题意：
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.

关键词：单向链表
思路：

1. 如何将前头接后尾？
2. 写一个findMid函数,遍历全链表，找到中点。用fast,slow方法
3. 写一个reverse函数，Reverse后半段
4. 将中点断开，形成连个子链表，前链表（包括中点） 和后链表（从中点下一个开始）。
5. 写一个merge(alist,blist),穿插接上，考虑用even,old, index

解答:

var reorderList = function (head) {
   if (!head) return;
   let mid = findMid(head);
   let second = reverse(mid.next);
   mid.next = null; //split two lists
   merge(head,second);
};

var findMid = function (head) {
   let slow = head;
   let fast = head;
   while (fast && fast.next) {
       slow = slow.next;
       fast = fast.next.next;
   }
   return slow;
}
var reverse = function(head){
   let pre = null;
   let cur = head;
   while (cur){
       let temp = cur.next; //before cut, store nextnode
       cur.next = pre;
       pre = cur;
       cur = temp; //make cycle
   }
   return pre;
}
var merge = function (l1, l2) {
   //merge two lists, need create new list /modify list need dummy
   let dummy = new ListNode(-1);
   //l1 include mid, l1'size > l2'size
   let index = 0;
   while (l2){
       if (index % 2 == 0){
           dummy.next = l1;
           l1 = l1.next;
       }
       else{
           dummy.next = l2;
           l2 = l2.next;
       }
       index++;
       dummy = dummy.next;
   }
   //because l2 < l1 add rest l1
   if (l1) dummy.next =l1;
}