用js开始刷力扣，坚持是第一位!!!!俾啲心机啊崽！

思路：

• 设置一个虚拟头结点再进行删除操作，因此创建新节点 ret const ret = new ListNode(0, head)
• return 头结点的时候 return cur.next

注意事项：
function ListNode的写法

function ListNode(val, next) {
    this.val = (val===undefined ? 0 : val)
    this.next = (next===undefined ? null : next)
}
 
var removeElements = function(head, val) {
    const ret = new ListNode(0, head)
    cur = ret;
    while(cur.next){
        if(cur.next.val === val){
            cur.next = cur.next.next;
        } else {
            cur = cur.next;
        }
    }
    return ret.next;
};

• 递归的方式

var removeElements = function (head, val) {
  if (head == null) {
    return head;
  }
  head.next = removeElements(head.next, val);
  // 只需要考虑当前节点是否等于val值，如果当前节点等于val值，则返回当前节点的下一个节点，否则返回当前节点
  return head.val == val ? head.next : head;
};

思路：

• 考察链表的基本逻辑
• 设置一个虚拟头结点
• 注释写在代码中，方便理解

注意事项：
无

class LinkNode {
    constructor(val, next) {
        this.val = val;
        this.next = next;
    }
}

/**
 * Initialize your data structure here.
 * 单链表 储存头尾节点 和 节点数量
 */
var MyLinkedList = function() {
    this._size = 0;
    this._tail = null;
    this._head = null;
};

/**
 * Get the value of the index-th node in the linked list. If the index is invalid, return -1. 
 * @param {number} index
 * @return {number}
 */
MyLinkedList.prototype.getNode = function(index) {
    if(index < 0 || index >= this._size) return null;
    // 创建虚拟头节点
    let cur = new LinkNode(0, this._head);
    // 0 -> head
    while(index-- >= 0) {
        cur = cur.next;
    }
    return cur;
};
MyLinkedList.prototype.get = function(index) {
    if(index < 0 || index >= this._size) return -1;
    // 获取当前节点
    return this.getNode(index).val;
};

/**
 * Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list. 
 * @param {number} val
 * @return {void}
 */
MyLinkedList.prototype.addAtHead = function(val) {
    const node = new LinkNode(val, this._head);
    this._head = node;
    this._size++;
    if(!this._tail) {
        this._tail = node;
    }
};

/**
 * Append a node of value val to the last element of the linked list. 
 * @param {number} val
 * @return {void}
 */
MyLinkedList.prototype.addAtTail = function(val) {
    const node = new LinkNode(val, null);
    this._size++;
    if(this._tail) {
        this._tail.next = node;
        this._tail = node;
        return;
    }
    this._tail = node;
    this._head = node;
};

/**
 * Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted. 
 * @param {number} index 
 * @param {number} val
 * @return {void}
 */
MyLinkedList.prototype.addAtIndex = function(index, val) {
    if(index > this._size) return;
    if(index <= 0) {
        this.addAtHead(val);
        return;
    }
    if(index === this._size) {
        this.addAtTail(val);
        return;
    }
    // 获取目标节点的上一个的节点
    const node = this.getNode(index - 1);
    node.next = new LinkNode(val, node.next);
    this._size++;
};

/**
 * Delete the index-th node in the linked list, if the index is valid. 
 * @param {number} index
 * @return {void}
 */
MyLinkedList.prototype.deleteAtIndex = function(index) {
    if(index < 0 || index >= this._size) return;
    if(index === 0) {
        this._head = this._head.next;
        // 如果删除的这个节点同时是尾节点，要处理尾节点
        if(index === this._size - 1){
            this._tail = this._head
        }
        this._size--;
        return;
    }
    // 获取目标节点的上一个的节点
    const node = this.getNode(index - 1);    
    node.next = node.next.next;
    // 处理尾节点
    if(index === this._size - 1) {
        this._tail = node;
    }
    this._size--;
};


/**
 * Your MyLinkedList object will be instantiated and called as such:
 * var obj = new MyLinkedList()
 * var param_1 = obj.get(index)
 * obj.addAtHead(val)
 * obj.addAtTail(val)
 * obj.addAtIndex(index,val)
 * obj.deleteAtIndex(index)
 */

思路：

• 在遍历链表时，将当前节点的 next 指针改为指向前一个节点。由于节点没有引用其前一个节点，因此必须事先存储其前一个节点。在更改引用之前，还需要存储后一个节点。最后返回新的头引用。

注意事项：
无

// 迭代：
var reverseList = function(head) {
    let prev = null;
    let curr = head;
    while (curr) {
        const next = curr.next;
        curr.next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
};

// 递归：
var reverseList = function(head) {
    if (head == null || head.next == null) {
        return head;
    }
    const newHead = reverseList(head.next);
    head.next.next = head;
    head.next = null;
    return newHead;
};

思路：

• 创建虚拟头节点
• 迭代

注意事项：
无

var swapPairs = function (head) {
  let ret = new ListNode(0, head), temp = ret;
  while (temp.next && temp.next.next) {
    let cur = temp.next.next, pre = temp.next;
    pre.next = cur.next;
    cur.next = pre;
    temp.next = cur;
    temp = pre;
  }
  return ret.next;
};