[LeetCode] Swap Nodes in Pairs 成對交換節點

 

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

Example:

Given 1->2->3->4, you should return the list as 2->1->4->3.

Note:

• Your algorithm should use only constant extra space.
• You may not modify the values in the list's nodes, only nodes itself may be changed.

 

這道題不算難，是基本的連結串列操作題，我們可以分別用遞迴和迭代來實現。對於迭代實現，還是需要建立dummy節點，注意在連線節點的時候，最好畫個圖，以免把自己搞暈了，參見程式碼如下：

 

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;
    }
};

 

遞迴的寫法就更簡潔了，實際上利用了回溯的思想，遞迴遍歷到連結串列末尾，然後先交換末尾兩個，然後依次往前交換：

 

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;
    }
};

 

類似題目：

Reverse Nodes in k-Group

 

參考資料：

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

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

 

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