题解 | 链表的奇偶重排

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    ListNode* oddEvenList(ListNode* head) {
        // write code here
        auto temp=head;
        auto first=new ListNode(0);
        auto second=new ListNode(0);
        ListNode* firstHead=first;
        ListNode* secondHead=second;
        int count=0;
        while(temp){
            count++;
            if(count%2){
                first->next=new ListNode(temp->val);
                first=first->next;
            }
            else{
                second->next=new ListNode(temp->val);
                second=second->next;
            }
            temp=temp->next;
        }
        //将第一个链表的头节点删除
        if(firstHead){
            temp=firstHead->next;
            firstHead->next=nullptr;
            delete firstHead;
            firstHead=temp;
        }
        //将第二个链表的头节点删除
        if(secondHead){
            temp=secondHead->next;
            secondHead->next=nullptr;
            delete secondHead;
            secondHead=temp;
        }
	  //将两个链表合并
        if(firstHead&&secondHead){
            first->next=secondHead;
        }
	  //打印
        printList(firstHead);
        printList(secondHead);
        return firstHead;
    }
    void printList(ListNode* head){
        while(head){
            std::cout<<head->val<<" ";
            head=head->next;
        }
        std::cout<<endl;
    }
};