力扣题解最佳实践  >  链表  >  206. 反转链表  >  已用 Java, Python, C++, JavaScript, C#, Go, Ruby 语言实现  >  转发 或 贡献代码

    力扣链接：206. 反转链表，难度：简单。

    给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。

    示例 1:

    输入: head = [1,2,3,4,5]

    输出: [5,4,3,2,1]

    示例 2:

    输入: [1,2]

    输出: [2,1]

    示例 3:

    输入: []

    输出: []

    约束:

    • 链表中节点的数目范围是 [0, 5000]
    • -5000 <= Node.val <= 5000

    思路
    上一题下一题

    1. 解决这个问题，只需要定义两个变量：current和previous。

      点击查看答案

      current.next = previous 就是反转了。

    2. 循环条件是while (current != null)，还是while (current.next != null)？

      点击查看答案

      是 while (current != null)，因为需要操作的是 current.next = previous。

    步骤

    1. 遍历所有节点。

      previous = null
current = head

while (current != null) {
    current = current.next
}

    2. 加入current.next = previous。

      previous = null
current = head

while (current != null) {
    tempNext = current.next
    current.next = previous
    current = tempNext
}

    3. previous目前始终是null，需要让它变化起来：previous = current。

      previous = null
current = head

while (current != null) {
    tempNext = current.next
    current.next = previous
    previous = current
    current = tempNext
}

    复杂度

    时间复杂度

    O(N)

    空间复杂度

    O(1)

    题解语言： Java Python C++ JavaScript C# Go Ruby

    Java # 

    /**
 * 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 previous = null;
        var current = head;

        while (current != null) {
            var tempNext = current.next;
            current.next = previous;
            previous = current;
            current = tempNext;
        }

        return previous;
    }
}

    Python # 

    # class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        previous = None
        current = head

        while current:
            temp_next = current.next
            current.next = previous
            previous = current
            current = temp_next

        return previous

    C++ # 

    /**
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* previous = nullptr;
        ListNode* current = head;

        while (current) {
            auto temp_next = current->next;
            current->next = previous;
            previous = current;
            current = temp_next;
        }

        return previous;
    }
};

    JavaScript # 

    /**
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
var reverseList = function (head) {
  let previous = null
  let current = head

  while (current != null) {
    const tempNext = current.next
    current.next = previous
    previous = current
    current = tempNext
  }

  return previous
};

    C# # 

    /**
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution
{
    public ListNode ReverseList(ListNode head)
    {
        ListNode previous = null;
        ListNode current = head;

        while (current != null)
        {
            var tempNext = current.next;
            current.next = previous;
            previous = current;
            current = tempNext;
        }

        return previous;
    }
}

    Go # 

    /**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
    var previous *ListNode
    current := head

    for current != nil {
        tempNext := current.Next
        current.Next = previous
        previous = current
        current = tempNext
    }

    return previous
}

    Ruby # 

    # class ListNode
#     attr_accessor :val, :next
# 
#     def initialize(val = 0, _next = nil)
#         @val = val
#         @next = _next
#     end
# end

def reverse_list(head)
  previous = nil
  current = head

  while current
    temp_next = current.next
    current.next = previous
    previous = current
    current = temp_next
  end

  previous
end
    题解语言： Java Python C++ JavaScript C# Go Ruby

    其它语言

    欢迎贡献代码到我们的 GitHub 仓库，非常感谢！本题解位置在 206. 反转链表。
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