[leetcode]143. Reorder List -- JavaScript 代码

/**
* 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===null || head.next===null){
return;
}
var top = head;
var mid = getMid(head);
var reversedlist = reverseList(mid);
mergeList(top,reversedlist);

function reverseList(firstNode){ // 1->2->3->4->5
if(firstNode===null || firstNode.next===null){
return firstNode;
}
var node1 = firstNode;
var node2 = firstNode.next;
node1.next = null;
while(node2.next!==null){
var tmp = node2.next;
node2.next = node1;
node1 = node2;
node2 = tmp;
}
node2.next = node1;
return node2;// the first node of the reversed list
}
function mergeList(l1,l2){// 1->2->3
if(l1===null){        // 4->5->6
return l2;
}
if(l2===null){
return l1;
}
while(l1!==null && l2!==null){
tmp = l1.next;
l1.next = l2;
l2 = l2.next;
l1.next.next = tmp;
l1 = tmp;
}
}
function getMid(first){ // 快慢指针获得中间点
var fast = first.next;
var slow = first.next;
var prev = first;
while(true)
{
if(fast !== null)
fast = fast.next;
else
break;
if(fast !== null)
fast = fast.next;
else
break;
prev = slow;
slow = slow.next;
}
prev.next = null;  // cut
return slow;
}
};

解题思路分为三部分：

1、function getMid：找到中间节点，将链表一分为二。

2、function reverseList：将第二个链表倒置。

3、function mergeList：将两个链表合并。

题目整体略微复杂，但是分开3个小问题，每个都比较好解决。