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LeetCode-092-Reverse Linked List II

雄狮虎豹
中文

Reverse Linked List II

Title description: Give you the head pointer of the singly linked list and two integers left and right, where left <= right. Please reverse the linked list node from position left to position right, and return to the reversed linked list.

Please refer to LeetCode official website for example description.

Source: LeetCode
Link: https://leetcode-cn.com/problems/reverse-linked-list-ii/
The copyright belongs to Lingkou Network. For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.

Solution 1: Use the stack

First, if head is null or head has only one node, return head directly.

Otherwise, declare a new head node newHead and declare a stack reverseNodes to put left and right for the node between the positions (for the reverse order):

  • Traverse the nodes in head
  • Put left into the new linked list once;
  • Put the nodes between left and right reverseNodes ;
  • Use rightNode record the position of the node after the right position;
  • Finally, put reverseNodes into the new linked list once, and then put rightNode at the end of the new linked list.

Finally, return newHead.next the linked list after inversion.

import com.kaesar.leetcode.ListNode;

import java.util.Stack;

public class LeetCode_092 {
    public static ListNode reverseBetween(ListNode head, int left, int right) {
        if (head == null || head.next == null) {
            return head;
        }
        // 声明一个新的头节点
        ListNode newHead = new ListNode(-1);
        ListNode leftNode = newHead, rightNode = head;
        // 记录是否已经走过left和right位置
        boolean findLeft = false, findRight = false;
        // 将left和right之间的节点放入栈中
        Stack<ListNode> reverseNodes = new Stack<>();
        int count = 1;
        while (head != null) {
            if (findLeft && findRight) {
                break;
            }
            if (findLeft) {
                if (count == right) {
                    reverseNodes.add(head);
                    rightNode = head.next;
                    break;
                } else {
                    reverseNodes.add(head);
                    head = head.next;
                }
            } else {
                if (count == left) {
                    findLeft = true;
                    reverseNodes.add(head);
                    if (count == right) {
                        rightNode = head.next;
                        findRight = true;
                        break;
                    }
                } else {
                    leftNode.next = head;
                    leftNode = leftNode.next;
                }
                head = head.next;
            }
            count++;
        }
        // 最后将栈中的节点逆序放入新的链表中
        while (!reverseNodes.isEmpty()) {
            leftNode.next = reverseNodes.pop();
            leftNode = leftNode.next;
        }
        leftNode.next = rightNode;
        return newHead.next;
    }

    public static void main(String[] args) {
        ListNode head = new ListNode(3);
        head.next = new ListNode(5);

        ListNode result = reverseBetween(head, 1, 2);
        while (result != null) {
            System.out.print(result.val + " ");
            result = result.next;
        }
    }
}
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