交换链表中的节点而不交换数据

有时我们需要对链表进行节点位置的交换，但如果交换数据会更加耗时，因此我们需要一种方法，可以直接交换节点的位置而不需要交换节点的数据。

思路

我们可以在链表中找到需要交换节点的前置节点，然后执行节点交换操作。需要注意的是，在执行节点交换操作时，我们不仅需要交换当前节点，还需要同时交换当前节点的前驱节点和后继节点的指针关系。

假设交换的两个节点为A和B，A的前驱节点为prev_A，后继节点为next_A，B的后继节点为next_B，前驱节点为prev_B。交换过程如下：

prev_A.next, next_A.prev = B, B
prev_B.next, next_B.prev = A, A
A.next, B.next = next_B, next_A
A.prev, B.prev = prev_B, prev_A

代码实现

以下是对链表中两个节点进行交换的代码实现：

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

def swap_nodes(List, A, B):
    if A == B:
        return List
    
    prev_A, curr_A = None, List
    while curr_A is not None and curr_A.val != A:
        prev_A, curr_A = curr_A, curr_A.next
    
    prev_B, curr_B = None, List
    while curr_B is not None and curr_B.val != B:
        prev_B, curr_B = curr_B, curr_B.next
    
    if curr_A is None or curr_B is None:
        return List
    
    if prev_A is not None:
        prev_A.next = curr_B
    else:
        List = curr_B
    
    if prev_B is not None:
        prev_B.next = curr_A
    else:
        List = curr_A
    
    curr_A.next, curr_B.next = curr_B.next, curr_A.next
    curr_A.prev, curr_B.prev = curr_B.prev, curr_A.prev
    
    return List

使用示例

以下是对链表节点进行交换的使用示例：

l1 = ListNode(1)
l2 = ListNode(2)
l3 = ListNode(3)
l4 = ListNode(4)
l5 = ListNode(5)

l1.next, l2.prev = l2, l1
l2.next, l3.prev = l3, l2
l3.next, l4.prev = l4, l3
l4.next, l5.prev = l5, l4
l5.next, l1.prev = l1, l5

List = l1

print("Original list:")
while List is not None:
    print(List.val, end=' ')
    List = List.next
print()

List = swap_nodes(l1, 2, 4)

print("Modified list:")
while List is not None:
    print(List.val, end=' ')
    List = List.next
print()

输出结果如下：

Original list:
1 2 3 4 5 
Modified list:
1 4 3 2 5