24. 两两交换链表中的节点 - 已题解语言：Java, Python, C++, JavaScript, C#, Go, Ruby

力扣题解最佳实践 > 链表 > 24. 两两交换链表中的节点 > 已用 Java, Python, C++, JavaScript, C#, Go, Ruby 语言实现 > 贡献代码或转发

力扣链接：24. 两两交换链表中的节点，难度：中等。

给你一个链表，两两交换其中相邻的节点，并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题（即，只能进行节点交换）。

示例 1:

输入: head = [1,2,3,4]

输出: [2,1,4,3]

示例 2:

输入: head = []

输出: []

示例 3:

输入: head = [1]

输出: [1]

示例 4:

输入: head = [1,2,3]

输出: [2,1,3]

约束:

链表中节点的数目在范围 [0, 100] 内
0 <= Node.val <= 100

思路
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在做本题前，建议先完成简单题目206. Reverse Linked List。

解决这个问题，依然至少需要定义两个变量：current和previous。
循环条件应是while (current.next != null)，而不应该是while (current != null)，因为需要操作current.next.next。

步骤

遍历所有节点。

var previous = null;
var current = head;

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

因为每两个节点进行一次位置互换，所以需要改为一次走两步。

var previous = null;
var current = head;

while (current != null && current.next != null) { // 1
    var nextNext = current.next.next; // 2
    current = nextNext; // 3
}

交换 current 和 current.next 的位置。

var previous = null;
var current = head;

while (current != null && current.next != null) {
    var nextNext = current.next.next;

    current.next.next = current; // 1
    current.next = nextNext; // 2

    current = nextNext;
}

处理 previous。

var previous = null;
var current = head;

while (current != null && current.next != null) {
    var nextNext = current.next.next;

    previous.next = current.next; // 1
    current.next.next = current;
    current.next = nextNext;

    previous = current; // 2
    current = nextNext;
}

确定返回值。因为head节点在节点数量超过1时，会被交换到第二个节点的位置，为了统一方便处理，最好加入dummy_head节点。

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

var previous = dummyHead; // 3
var current = head;

while (current != null && current.next != null) {
    var nextNext = current.next.next;

    previous.next = current.next;
    current.next.next = current;
    current.next = nextNext;

    previous = current;
    current = nextNext;
}

return dummyHead.next; // 4

复杂度

时间复杂度

O(N)

空间复杂度

O(1)

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 swapPairs(ListNode head) {
        var dummyHead = new ListNode(0, head);
        var previous = dummyHead;
        var current = head;

        while (current != null && current.next != null) {
            var nextNext = current.next.next;

            previous.next = current.next;
            current.next.next = current;
            current.next = nextNext;

            previous = current;
            current = nextNext;
        }

        return dummyHead.next;
    }
}

Python #

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

class Solution:
    def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy_head = ListNode(next=head)
        previous = dummy_head
        current = head

        while current and current.next:
            next_next = current.next.next

            previous.next = current.next
            current.next.next = current
            current.next = next_next

            previous = current
            current = next_next

        return dummy_head.next

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* swapPairs(ListNode* head) {
        auto dummy_head = new ListNode(0, head);
        auto previous = dummy_head;
        auto current = head;

        while (current != nullptr && current->next != nullptr) {
            auto next_next = current->next->next;

            previous->next = current->next;
            current->next->next = current;
            current->next = next_next;

            previous = current;
            current = next_next;
        }

        auto result = dummy_head->next;
        delete dummy_head;
        return result;
    }
};

JavaScript #

/**
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
var swapPairs = function (head) {
  const dummyHead = new ListNode(0, head)

  let previous = dummyHead
  let current = head

  while (current != null && current.next != null) {
    const nextNext = current.next.next

    previous.next = current.next
    current.next.next = current
    current.next = nextNext

    previous = current
    current = nextNext
  }

  return dummyHead.next
};

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 SwapPairs(ListNode head)
    {
        var dummyHead = new ListNode(0, head);
        var previous = dummyHead;
        var current = head;

        while (current != null && current.next != null)
        {
            var nextNext = current.next.next;

            previous.next = current.next;
            current.next.next = current;
            current.next = nextNext;

            previous = current;
            current = nextNext;
        }

        return dummyHead.next;
    }
}

Go #

/**
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func swapPairs(head *ListNode) *ListNode {
    dummyHead := &ListNode{0, head}

    previous := dummyHead
    current := head

    for current != nil && current.Next != nil {
        nextNext := current.Next.Next

        previous.Next = current.Next
        current.Next.Next = current
        current.Next = nextNext

        previous = current
        current = nextNext
    }

    return dummyHead.Next
}

Ruby #

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

def swap_pairs(head)
  dummy_head = ListNode.new
  dummy_head.next = head

  previous = dummy_head
  current = head

  while current && current.next
    next_next = current.next.next

    previous.next = current.next
    current.next.next = current
    current.next = next_next

    previous = current
    current = next_next
  end

  dummy_head.next
end

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力扣题解最佳实践 > 链表 > 24. 两两交换链表中的节点 > 已用 Java, Python, C++, JavaScript, C#, Go, Ruby 语言实现 > 贡献代码 或 转发