C++实现LeetCode(24.成对交换节点)
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2022-07-07 23:34:38
[leetcode] 24. swap nodes in pairs 成对交换节点given a linked list, swap every two adjacent nodes and...
[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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