24. Swap Nodes in Pairs

 

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;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/24

 

类似题目:

Reverse Nodes in k-Group

 

参考资料:

https://leetcode.com/problems/swap-nodes-in-pairs

https://leetcode.com/problems/swap-nodes-in-pairs/discuss/11030/My-accepted-java-code.-used-recursion.

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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