考研算法

题目

题目链接

题目要求

合并链表+反转列表

C++代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void rearrangedList(ListNode* head) {
        if (head->next==NULL) return ;
        int len=0;
        auto p=head;
        while(p)
        {
            len++;
            p=p->next;
        }
        if (len==1)
        {
            return ;
        }
        int mid=(len+1)/2;
        auto a=head;
        for (int i=0;i<mid-1;i++)
        {
            a=a->next;
        }
        auto b=a->next;
        auto c=b->next;
        a->next=NULL;
        b->next=NULL;
        while(c)
        {
            auto t=c->next;
            c->next=b;
            b=c;
            c=t;
        }
        for (auto p=head,q=b;q;)
        {
            auto temp=q->next;
            q->next=p->next;
            p->next=q;
            p=p->next->next;
            q=temp;
        }
    }
};

知识点

先将链表从中间分为左右两个链表，再将右链表反转，最后将两个链表合并
反转列表

    int mid=(len+1)/2;
    auto a=head;
    for (int i=0;i<mid-1;i++)//寻找左链表结束点
    {
        a=a->next;
    }
    auto b=a->next;
    auto c=b->next;
    a->next=NULL;//需要设置为null，让它断开连接，否则最后的循环无法退出
    b->next=NULL;//需要设置为null，让它断开连接，否则最后的循环无法退出
    while(c)//反转
    {
        auto t=c->next;
        c->next=b;
        b=c;
        c=t;
    }

合并链表

    for (auto p=head,q=b;q;)//合并
    {
        auto temp=q->next;
        q->next=p->next;
        p->next=q;
        p=p->next->next;
        q=temp;
    }

python代码

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def rearrangedList(self, node):
        """
        :type node: ListNode
        :rtype: void 
        """
        if node.next==None :return 0
        le=0;
        p=node
        while(p):
            p=p.next
            le=le+1
        mid=(le+1)//2
        a=node
        for i in range(mid-1):
            a=a.next
        b=a.next
        c=b.next
        a.next=None
        b.next=None
        while(c):#反转
            p=c.next
            c.next=b
            b=c
            c=p
        x=node
        z=b
        while(z): #合并
            t=z.next
            z.next=x.next
            x.next=z
            x=x.next.next
            z=t