LeetCode data structure and algorithm reverse linked list

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topic

Reverse Linked List Give you the head node of the singly linked list, please reverse the linked list and return the reversed linked list. Example 1:

输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]
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Example 2:

输入：head = [1,2]
输出：[2,1]
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Example 3:

输入：head = []
输出：[]
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hint:

• The number of nodes in the linked list ranges from[0, 5000]
• -5000 <= Node.val <= 5000

Advanced: The linked list can be reversed in an iterative or recursive way. Can you solve this problem in two ways?

answer

problem-solving analysis

Problem solving ideas

• Iterative processing

1. Suppose the linked list data is 1->2->3->end, we need to modify it to end->3->2-1.
2. When traversing, modify the current nextpointer to point to the previous node. Since the node does not refer to the previous node, the previous node must be stored beforehand. Before modifying the reference, it is also necessary to store the latter node, and return the new head node after the final execution is completed.
3. It takes one cycle to complete the reversal.

the complexity

Time complexity O(N)
Space complexity O(1)

problem solving code

The solution code is as follows (detailed comments in the code):

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode prev = null ;
        ListNode curr = head;
        // 直接迭代反转
        while(curr != null) {
            ListNode next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }
}
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Feedback results after submission (because this topic has not been optimized, the performance is average):

• Another solution (recursion), the core code is ead.next.next = head.

class Solution {
    public ListNode reverseList(ListNode head) {
        // 如果头为空，或者只有一个节点
        if (head == null || head.next == null) {
            // 递归出口
            return head;
        }
        // 递归调用
        ListNode newHead = reverseList(head.next);
        // 这里很难理解本质就是将
        // 1->2->end 转换为 end->2->1
        head.next.next = head;
        // 防止出现环
        head.next = null;
        return newHead;
    }
}
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Reference Information

• 206. Reverse linked list