C++实现LeetCode(24.成对交换节点)
[LeetCode] 24. Swap Nodes in Pairs 成对交换节点
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.
Example:
Given
1->2->3->4
, you should return the list as
2->1->4->3.
这道题不算难,是基本的链表操作题,我们可以分别用递归和迭代来实现。对于迭代实现,还是需要建立 dummy 节点,注意在连接节点的时候,最好画个图,以免把自己搞晕了,参见代码如下:
解法一:
class Solution { public: ListNode* swapPairs(ListNode* head) { ListNode *dummy = new ListNode(-1), *pre = dummy; dummy->next = head; while (pre->next && pre->next->next) { ListNode *t = pre->next->next; pre->next->next = t->next; t->next = pre->next; pre->next = t; pre = t->next; } return dummy->next; } };
递归的写法就更简洁了,实际上利用了回溯的思想,递归遍历到链表末尾,然后先交换末尾两个,然后依次往前交换:
解法二:
class Solution { public: ListNode* swapPairs(ListNode* head) { if (!head || !head->next) return head; ListNode *t = head->next; head->next = swapPairs(head->next->next); t->next = head; return t; } };
解法三:
class Solution { public ListNode swapPairs(ListNode head) { if (head == null || head.next == null) { return head; } ListNode newHead = head.next; head.next = swapPairs(newHead.next); newHead.next = head; return newHead; } }
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