LeetCode One Question a Day (39) 24. Swap nodes in the linked list in pairs

24. Pairwise exchange of nodes in the linked list

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answer:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
    
    
public:
    // 1. start->p1->p2->end
    // 2. p1和p2表示要交换的两个节点，我们需要借助start和end两个节点来辅助完成交换；
    ListNode* swapPairs(ListNode* head) {
    
    
        ListNode *start,*p1,*p2,*end;
        p1=head;
        start=head;
        // 每两个节点作为一对处理
        while(p1!=nullptr){
    
    
            if(p1->next==nullptr) break;
            // 情况1：head节点需要特殊处理，因为这种情况p1没有前驱节点
            if(start==head){
    
     
                p2=p1->next;
                end=p2->next;
                p1->next=end;
                p2->next=p1;
                head=p2;
            }
            // 情况2：需要交换的两个节点在链表的中间部分
            else{
    
    
                p2=p1->next;
                end=p2->next;
                p1->next=end;
                p2->next=p1;
                start->next=p2;
            }
            // 下一对节点
            start=p1;
            p1=p1->next;
        }
        return head;
    }
};

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Recursion:

class Solution {
    
    
public:
    ListNode* swapPairs(ListNode* head) {
    
    
        if (head == nullptr || head->next == nullptr) {
    
    
            return head;
        }
        ListNode* newHead = head->next;
        head->next = swapPairs(newHead->next);
        newHead->next = head;
        return newHead;
    }
};

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