题解 | #链表中的节点每k个一组翻转#

/*
 * 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
    if (!head || !head.next || k === 1) {
        return head;
    }
    let cur = head;
    let i = 0;
    let group = [];
    let curList;
    let count = 0;
    while (cur) {
        cur = cur.next;
        count++;
    }
    if (count < k) {
        return head;
    }
    cur = head;
    while (cur) {
        let temp = cur;
        cur = cur.next;
        if (i % k === 0) {
            if (i > 0) {
                group.push(curList);
            }
            if (k * (group.length + 1) > count) {
                group.push(temp);
                break;
            } else {
                curList = temp;
                curList.next = null;
            }
        } else {
            temp.next = curList;
            curList = temp;
        }
        i++;
    }
    if (count === k) {
        return curList;
    }
    head = group[0];
    for (let j = 0; j < group.length; j++) {
        let next = group[j];
        while (next) {
            if (!next.next) {
                next.next = group[j + 1];
                next = null;
            } else {
                next = next.next;
            }
        }
    }

    return head;
}
module.exports = {
    reverseKGroup: reverseKGroup,
};

解题思路：按组分别翻转，然后重新连接