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算法笔记: 力扣#206 反转链表

admin 11月 12, 2020 0

问题描述

解法

分析

Python 实现

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class :
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        prev = None
        cur = head
        while cur:
            temp = cur.next
            cur.next = prev
            prev = cur
            cur = temp
        return prev

Java 实现

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class  {
    public ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode cur = head;
        while(cur != null){
            ListNode temp = cur.next;
            cur.next = prev;
            prev = cur;
            cur = temp;
        }
        return prev;
    }
}

时间复杂度

O(n).

空间复杂度

O(1).

链接

206. Reverse Linked List
206. 反转链表
 (English version) Algorithm Notes: Leetcode#206 Reverse Linked List