【LeetCode】024. Swap Nodes in Pairs

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

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

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

题解：

　　没什么好写的，链表题一般要注意头结点的问题，还有一定要画图！

Solution 1

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode dummy(-1);
        dummy.next = head;
        ListNode* cur = &dummy;
        
        while (cur->next && cur->next->next) {
            ListNode* tmp = cur->next->next;
            cur->next->next = tmp->next;
            tmp->next = cur->next;
            cur->next = tmp;
            cur = tmp->next;
        }
        
        return dummy.next;
    }
};

　　递归：

Solution 2

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        if (!head || !head->next)
            return head;
        ListNode* newHead = head->next;
        head->next = swapPairs(head->next->next);
        newHead->next = head;
        return newHead;
    }
};