问题：交换链表中的两个节点：

Question

这有两种可能的情况：

1. Node_1 和 Node_2 不是头节点
2. Node_1 或 Node_2 都是头节点

代码：

def swap_nodes(self, key_1, key_2):
    if key_1 == key_2:
        return
    
    prev_1 = None
    curr_1 = self.head
    while curr_1 and curr_1.data != key_1:
        prev_1 = curr_1
        curr_1 = curr_1.next
        
    prev_2 = None
    curr_2 = self.head
    while curr_2 and curr_2.data != key_1:
        prev_2 = curr_2
        curr_2 = curr_2.next
        
    if not curr_1 or not curr_2:
        return
    
    if prev_1:                            #doubt from here 
        prev_1.next = curr_2              
    else:                               
        self.head = curr_2                
        
    if prev_2:
        prev_2.next = curr_1
    else:
        self.head = curr_1                #doubt till here
        
    curr_1.next, curr_2.next = curr_2.next, curr_1.next

我无法理解这部分提到的疑问。有人可以解释这是如何工作的

Answer 1

在单链表中（一个 next 指针，但没有 prev 指针）当你想删除时你必须记住前一个节点或移动一个节点。

这是您的代码的注释版本：

def swap_nodes(self, key_1, key_2):
    """Swap nodes containing resp. key1 and key2 values in the linked list starting at head"""
    if key_1 == key_2:           # same values: nothing to do
        return
    
    # search key1 and remember its previous node (prev1 is None <=> key1 in head node)
    prev_1 = None
    curr_1 = self.head
    while curr_1 and curr_1.data != key_1:
        prev_1 = curr_1
        curr_1 = curr_1.next
        
    # same for key2
    prev_2 = None
    curr_2 = self.head
    while curr_2 and curr_2.data != key_1:
        prev_2 = curr_2
        curr_2 = curr_2.next
        
    if not curr_1 or not curr_2:     # one key not found
        return
    
    # make previous nodes of curr1 and curr2 point to the other ("back" links)
    if prev_1:                            # is key1 n head node ?
        prev_1.next = curr_2              # prev1 must point to curr2
    else:                               
        self.head = curr_2                # new head will be curr2
        
    if prev_2:                            # same for second node
        prev_2.next = curr_1
    else:
        self.head = curr_1                #doubt till here
        
    # "back" links are done, now for the direct links (simpler isn't it?)
    curr_1.next, curr_2.next = curr_2.next, curr_1.next

希望现在更清楚...