234. 回文链表

最新推荐文章于 2024-01-19 21:57:04 发布

原创最新推荐文章于 2024-01-19 21:57:04 发布 · 91 阅读

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这个博客主要介绍了如何使用Java实现链表的各种操作，包括判断链表是否为回文、在指定位置反转链表、分区链表以及按指定步长旋转链表。这些操作均在O(n)时间复杂度和O(1)空间复杂度下完成，展示了高效的数据结构处理能力。

请判断一个链表是否为回文链表。

示例 1:

输入: 1->2
输出: false

示例 2:

输入: 1->2->2->1
输出: true

进阶：
你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题？

package Solution234;

import java.util.ArrayList;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.Stack;

class Solution {
	public boolean isPalindrome(ListNode head) {
		Stack<ListNode> stk = new Stack<>();
		ListNode cur = head;
		while (cur != null) {// 放入stack
			stk.push(cur);
			cur = cur.next;
		}
		while (head != null && head.next != null) {

			if (head.val != stk.pop().val) {
				return false;
			}

			head = head.next;
		}
		return true;
	}

	public ListNode reverseBetween(ListNode head, int left, int right) {
		Stack<ListNode> stk = new Stack<ListNode>();

		if (head == null || head.next == null)
			return head;
		ListNode dummy1 = new ListNode(-1);
		ListNode s = dummy1;

		ListNode dummy2 = new ListNode(-1);
		ListNode b = dummy2;

		ListNode curr = head;
		int count = 1;
		while (curr != null) {
			if (count < left) {
				s.next = curr;
				s = s.next;
			}
			if (count > right) {
				b.next = curr;
				b = b.next;
			}
			if (count >= left && count <= right) {
				stk.push(curr);

			}

			curr = curr.next;
			count++;
		}
		b.next = null;
		while (!stk.isEmpty()) {
			s.next = stk.pop();
			s = s.next;
		}

		s.next = dummy2.next;
		return dummy1.next;

	}

	public ListNode partition(ListNode head, int x) {
		if (head == null || head.next == null)
			return head;
		ListNode dummy1 = new ListNode(-1);
		ListNode s = dummy1;

		ListNode dummy2 = new ListNode(-1);
		ListNode b = dummy2;

		ListNode curr = head;
		while (curr != null) {
			if (curr.val < x) {
				s.next = curr;
				s = s.next;
			} else {
				b.next = curr;
				b = b.next;
			}
			curr = curr.next;
		}
		b.next = null;
		s.next = dummy2.next;
		return dummy1.next;
	}

	public ListNode rotateRight(ListNode head, int k) {
		if (head == null) {
			return head;
		}
		if (k == 0) {
			return head;
		}

		int count = 0;
		ListNode start = head;
		// traverse till the end of the list and keep incrementing
		// count
		while (head.next != null) {
			head = head.next;
			count++;
		}
		count++;
		// if k > count, do k%count, its an optimization. 2%5 == 12%5
		k = k % count;
		// find the new k
		k = Math.abs(count - k);
		if (k == 0)
			return start;
		// connect last element to start
		head.next = start;
		// traverse for new k value
		while (k-- > 0) {
			head = head.next;
		}
		// note: head is not the last element, so set the start.
		start = head.next;
		head.next = null;
		return start;
	}

	public ListNode mergeKLists(ListNode[] lists) {
		if (lists == null || lists.length == 0)
			return null;

		PriorityQueue<ListNode> queue = new PriorityQueue<ListNode>(
				(Comparator<? super ListNode>) new Comparator<ListNode>() {
					public int compare(ListNode l1, ListNode l2) {
						return l1.val - l2.val;
					}
				});

		ListNode head = new ListNode(0);
		ListNode p = head;

		for (ListNode list : lists) {
			if (list != null)
				queue.offer(list);
		}

		while (!queue.isEmpty()) {
			ListNode n = queue.poll();
			p.next = n;
			p = p.next;

			if (n.next != null)
				queue.offer(n.next);
		}

		return head.next;

	}

	public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
		if (l1 == null)
			return l2;
		if (l2 == null)
			return l1;
		if (l1.val < l2.val) {
			l1.next = mergeTwoLists(l1.next, l2);
			return l1;
		} else {
			l2.next = mergeTwoLists(l1, l2.next);
			return l2;
		}
	}

	public ListNode removeElements(ListNode head, int val) {
		ListNode header = new ListNode(-1);
		header.next = head;
		ListNode cur = header;
		while (cur.next != null) {
			if (cur.next.val == val) {
				cur.next = cur.next.next;
			} else {
				cur = cur.next;
			}
		}
		return header.next;
	}

	void printList(ListNode node) {
		while (node != null) {
			System.out.print(node.val + " ");
			node = node.next;
		}
	}

	public static void main(String[] args) {

		Solution sol = new Solution();

		ListNode head = new ListNode(1);
		head.next = new ListNode(2);
		head.next.next = new ListNode(3);
		head.next.next.next = new ListNode(2);
		head.next.next.next.next = new ListNode(1);
		// head.next.next.next.next.next = new ListNode(2);

		System.out.println(sol.isPalindrome(head));
		sol.printList(head);

	}

}