题解 | #牛群的重新分组#

/*
 * function ListNode(x){
 *   this.val = x;
 *   this.next = null;
 * }
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param head ListNode类 
 * @param k int整型 
 * @return ListNode类
 */
function reverseKGroup( head ,  k ) {
    // write code here
    let dummy = new ListNode(0)
    dummy.next = head
    let pre = dummy
    let cur = head
    let count = 0
    while(cur!=null){
        count++
        if(count%k==0){
            pre = reverse(pre,cur.next)
            cur = pre.next
        }else{
            cur = cur.next
        }
    }
    return dummy.next
    function reverse(pre,end){
        let cur = pre.next
        let tail = cur
        while(cur!=end){
            let temp = cur.next      
            cur.next = pre.next      //将链表依次逆转    
            pre.next = cur           //保存反转链表的前驱结点
            cur = temp
        }
        tail.next = end              //逆转后的最后一个结点与end结点连接
        console.log(tail)
        return tail
    }
}
module.exports = {
    reverseKGroup : reverseKGroup
};