Leetcode - 24 - Swap Nodes in Pairs-优快云博客

博客围绕链表展开，要求交换链表中每两个相邻节点并返回头节点，且不能修改节点的值，仅能改变节点本身。给出了递归和迭代两种解决方案，并分别展示了Java、Python、C++等语言的实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a linked list, swap every two adjacent nodes and return its head.

You may not modify the values in the list's nodes. Only nodes itself may be changed.

Solutions:

(1) Recursive: when head.next.next pairs are swapped, only need to swap the first two nodes

Java:

class Solution {
    public ListNode swapPairs(ListNode head) {
        if ((head == null) || (head.next == null)) 
            return head;
        ListNode q = swapPairs(head.next.next);
        ListNode r = head.next;
        r.next = head;
        head.next = q;
        return r;
    }
}

// OR
class Solution {
    public ListNode swapPairs(ListNode head) {
        if ((head == null) || (head.next == null)) 
            return head;
        ListNode r = head.next;
        head.next = swapPairs(head.next.next);
        r.next = head;
        return r;
    }
}

Python:

class Solution(object):
    def swapPairs(self, head):
        if not head or not head.next: return head
        new_start = head.next.next
        head, head.next = head.next, head
        head.next.next = self.swapPairs(new_start)
        return head

// https://leetcode.com/problems/swap-nodes-in-pairs/discuss/11046/My-simple-JAVA-solution-for-share

(2) Iterative:

C++:

// Pointer-pointer pp points to the pointer to the current node. So at first, pp points to head, and later it points to the next field of ListNodes.
//Additionally, for convenience and clarity, pointers a and b point to the current node and the next node.
//We need to go from *pp == a -> b -> (b->next) to *pp == b -> a -> (b->next). 
//The first three lines inside the loop do that, setting those three pointers (from right to left). The fourth line moves pp to the next pair.

ListNode* swapPairs(ListNode* head) {
    ListNode **pp = &head, *a, *b;
    while ((a = *pp) && (b = a->next)) {
        a->next = b->next;
        b->next = a;
        *pp = b;
        pp = &(a->next);
    }
    return head;
}
// https://leetcode.com/problems/swap-nodes-in-pairs/discuss/11019/7-8-lines-C%2B%2B-Python-Ruby

Python:

'''
Here, pre is the previous node. Since the head doesn't have a previous node, I just use self instead. Again, a is the current node and b is the next node.

To go from pre -> a -> b -> b.next to pre -> b -> a -> b.next, we need to change those three references. Instead of thinking about in what order I change them, I just change all three at once.
'''

def swapPairs(self, head):
    pre, pre.next = self, head
    while pre.next and pre.next.next:
        a = pre.next
        b = a.next
        pre.next, b.next, a.next = b, a, b.next
        pre = a
    return self.next

# https://leetcode.com/problems/swap-nodes-in-pairs/discuss/11019/7-8-lines-C%2B%2B-Python-Ruby

# use 'dummydummy = pre = ListNode(0)' instead of 'self'
class Solution:
    def swapPairs(self, head):
        dummy = pre = ListNode(0)
        pre.next = head
        while pre.next and pre.next.next:
            a = pre.next
            b = a.next
            pre.next, a.next, b.next = b, b.next, a
            pre = a
        return dummy.next

# by sfdye https://leetcode.com/problems/swap-nodes-in-pairs/discuss/11019/7-8-lines-C%2B%2B-Python-Ruby

Java:

public ListNode swapPairs(ListNode head) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode current = dummy;
    while (current.next != null && current.next.next != null) {
        ListNode first = current.next;
        ListNode second = current.next.next;
        first.next = second.next;
        current.next = second;
        current.next.next = first;
        current = current.next.next;
    }
    return dummy.next;
}

// https://leetcode.com/problems/swap-nodes-in-pairs/discuss/11046/My-simple-JAVA-solution-for-share