反转链表middle

eg：

输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]
相关解法：
图解：

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} left
 * @param {number} right
 * @return {ListNode}
 */
var reverseBetween = function(head, left, right) {
    let prev = null
    let cur = head
    let next = head
    for(let i=1;i<left;i++){ //从1 开始
        prev  = cur
        cur = cur.next
    }

    // 保留prev和cur
    let prev2 = prev
    let cur2 = cur
    //反转链表
    for(let i=left;i<=right;i++){
        next = cur.next
        cur.next = prev
        prev = cur
        cur = next
    }
    //判断是否left为第一个开始
    if(prev2 !== null){  //不是第一个
        prev2.next = prev
    }else{ //是第一个  prev为反转后的第一个结点
        head = prev
    }
    cur2.next = cur //反转后的最后一个结点指向cur

    return head
};

反转链表一：easy解法：

function ReverseList(pHead)
{
    // write code here
    let prev = null
    let cur = pHead
    let next = pHead
    while(cur){
        next = cur.next
        cur.next = prev
        prev = cur
        cur = next
    }
    return prev //此时cur指向null next也为null
}