Leetcode题解-24. Swap Nodes in Pairs

Leetcode题解-24. 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.

思路

仔细看这道题，其实只是第一二个节点、第三四个节点交换节点。
思路1：交换节点只进行交换值的操作，不对链表进行什么操作，只需要两个额外指针。
思路2：交换节点就是真的对链表进行操作，因此需要三个额外的指针进行标记。
需要注意的点是判断节点是否为空，防止对空指针进行操作

代码

思路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* first = head;
        if(!head) return nullptr;
        ListNode* second = head->next;
        while(second){     
            if(!first) break;
            int temp = first->val;
            first->val = second->val;
            second->val = temp;
            first = second->next;
            if(first) second = first->next;
        }
        return head;
    }
};

思路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) {
        ListNode pre(-1); pre.next = head;
        ListNode* prenode = ⪯
        ListNode* first = head;
        if(!head) return nullptr;
        ListNode* second = head->next;
        while(second){     
            if(!first) break;

            first->next = second->next;
            second->next = first;
            prenode->next = second;
            prenode = first;
            first = first->next;
            if(first) second = first->next;
        }
        return pre.next;
    }
};