0025. Reverse Nodes in k-Group (H)

题目

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list’s nodes, only nodes itself may be changed.

题意

将给定链表中的每k个结点按逆序重新排列，不能改变结点内的值，只能改动结点本身，且只能使用的空间。

思路

整体上与 0024. Swap Nodes in Pairs (M) 的解决方法相似，只是多了2个以上结点的逆序排列，因为只能使用常数的额外空间，所以不能用栈来实现逆序，直接使用插入到头结点之前的方式来实现逆序。

代码实现

Java

递归

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode pointer = head;
        // 判断剩余结点数是否达到k，并找到下一个k元组的头结点
        for (int i = 0; i < k; i++) {
            if (pointer == null) {
                return head;
            }
            pointer = pointer.next;
        }
        ListNode nextHead = pointer;        // 记录下一个k元组的头结点
        ListNode first = null;                // 逆序排列时记录逆序链表的头结点
        pointer = head;
        // 逆序处理
        while (pointer != nextHead) {
            ListNode temp = pointer.next;
            pointer.next = first;
            first = pointer;
            pointer = temp;
        }
        head.next = reverseKGroup(nextHead, k);
        return first;
    }
}

迭代

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode ans = new ListNode(0);
        ans.next = head;
        ListNode root = ans;        // 记录上一个k元组逆序后的最后一个结点
        // 统计结点数
        ListNode pointer = head;
        int num = 0;
        while (pointer != null) {
            num++;
            pointer = pointer.next;
        }
        pointer = ans.next;
        while (num >= k) {
            ListNode first = null;            // 逆序时的头结点
            ListNode nextRoot = pointer;    // 记录当前k元组逆序前的第一个结点
            // 逆序处理
            for (int i = 0; i < k; i++) {
                ListNode temp = pointer.next;
                pointer.next = first;
                first = pointer;
                pointer = temp;
            }
            root.next = first;
            root = nextRoot;
            num -= k;
        }
        root.next = pointer;        // 将个数不足k的结点补充到链表最后
        return ans.next;
    }
}

JavaScript

递归

/**
 * 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) {
  if (head === null) {
    return null
  }
  let nextHead = head
  for (let i = 1; i <= k; i++) {
    if (nextHead === null) {
      return head
    }
    nextHead = nextHead.next
  }
  let newHead = null
  let tail = head
  while (head !== nextHead) {
    let tmp = head.next
    head.next = newHead
    newHead = head
    head = tmp
  }
  tail.next = reverseKGroup(nextHead, k)
  return newHead
}

迭代

/**
 * 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 len = 0
  let p = head
  while (p !== null) {
    len++
    p = p.next
  }

  let dummy = new ListNode(0, head)
  let tail = dummy
  p = head
  while (len >= k) {
    let nextTail = p
    let newHead = null
    for (let i = 0; i < k; i++) {
      let tmp = p.next
      p.next = newHead
      newHead = p
      p = tmp
    }
    tail.next = newHead
    tail = nextTail
    len -= k
  }
  tail.next = p

  return dummy.next
}