leetcode 25 K 个一组翻转链表

题目

java 递归：

ListNode reverseKGroup(ListNode head, int k) {
    if (head == null) return null;
    // 区间 [a, b) 包含 k 个待反转元素
    ListNode a, b;
    a = b = head;
    for (int i = 0; i < k; i++) {
        // 不足 k 个，不需要反转，base case
        if (b == null) return head;
        b = b.next;
    }
    // 反转前 k 个元素
    ListNode newHead = reverse(a, b);
    // 递归反转后续链表并连接起来
    a.next = reverseKGroup(b, k);
    return newHead;
}

/** 反转区间 [a, b) 的元素，注意是左闭右开 */
ListNode reverse(ListNode a, ListNode b) {
    ListNode pre, cur, nxt;
    pre = null; cur = a; nxt = a;
    // while 终止的条件改一下就行了
    while (cur != b) {
        nxt = cur.next;
        cur.next = pre;
        pre = cur;
        cur = nxt;
    }
    // 返回反转后的头结点
    return pre;
}

java 栈：

/**
 * 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) {
        Deque<ListNode> stack = new ArrayDeque<ListNode>();
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        while (true) {
            int count = 0;
            ListNode tmp = head;
            while (tmp != null && count < k) {
                stack.add(tmp);
                tmp = tmp.next;
                count++;
            }
            if (count != k) {
                p.next = head;
                break;
            }
            while (!stack.isEmpty()){
                p.next = stack.pollLast();
                p = p.next;
            }
            p.next = tmp;
            head = tmp;
        }
        return dummy.next;
    }
}

java 尾插法：

/**
 * 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 dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        ListNode tail = dummy;
        while (true) {
            int count = 0;
            while (tail != null && count != k) {
                count++;
                tail = tail.next;
            }
            if (tail == null) break;
            ListNode head1 = pre.next;
            while (pre.next != tail) {
                ListNode cur = pre.next;
                pre.next = cur.next;
                cur.next = tail.next;
                tail.next = cur;
            }
            pre = head1;
            tail = head1;
        }
        return dummy.next;
    }
}

递归思路：

链表是一种兼具递归和迭代性质的数据结构
递归性质
子问题