25. K 个一组翻转链表 about algorithm HOT 2 OPEN

```js
/**
 * 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} k
 * @return {ListNode}
 */
var reverseKGroup = function(head, k) {
    let tmp = head;
  // 递归的出口就是当剩余的节点不满足k个时，直接返回当前链表的头节点
    for(let i = 0; i < k; i++){
        if(tmp==null) return head;
        tmp = tmp.next;
    }
   // 反转[a,b)
    const last = reverse(head,tmp);
    head.next = reverseKGroup(tmp,k);
    return last;
};

// 这里将反转链表更改一下，翻转特定两个节点之间的链表
// [a,b) 不包含右侧节点。
function reverse(head,b){
    let prev = null;
    let cur = head;
    while(cur!==b){
        const next = cur.next;
        cur.next = prev;
        prev = cur;
        cur = next;
    }
    return prev;
}

/**
 * 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} k
 * @return {ListNode}
 */
var reverseKGroup = function(head, k) {
    let tmp = head;
  // 这里我们只需要构建满足k个一组的链表即可，所以是k-1
    for(let i = 0; i < k - 1; i++){
        if(tmp==null) return head;
        tmp = tmp.next;
    }
    // 构建新链表
    let next = null;
    if(tmp!==null){
        next = tmp.next
        tmp.next = null;
    }else{
       // 如果当前tmp为null 证明不满足k个一组的需求
        return head;
    }
    
    const last = reverse(head);
    head.next = reverseKGroup(next,k);
    return last;
};


function reverse(head){
    if(head==null||head.next==null)return head;
    const last = reverse(head.next);
    head.next.next =head;
    head.next = null;
    return last;
}