高频链表面试题

#include<iostream>
using namespace std;
#include<list>
struct ListNode{
    int data;
    struct ListNode* next;
    ListNode(int x):data(x),next(NULL)
    {}
};
//删除不带头的非尾节点
void delNodeTail(ListNode* pos){
    assert(pos&&pos->next);
        ListNode* next = new ListNode(0);
        next = pos->next;
        pos->data = next ->data;
        pos->next = next->next;
        free(next);
}
//从尾到头的打印链表
void printTailtoHead(ListNode* head){
    if(head==NULL)
      return;
    printTailtoHead(head->next);
    printf("%d",head->data);
}
//在无头的单链表前面插入一个节点,思路：新建一个pos位置的节点，将原pos位置的节点的值置为x，再将新pos的位置插入即可
void insertNodeFront(ListNode* pos,int x){
    assert(pos);
    ListNode* node = new ListNode(pos->data);
    ListNode* next = pos->next;
    pos->data = x;
void insertNodeFront(ListNode* pos,int x){
    assert(pos);
    ListNode* node = new ListNode(pos->data);
    ListNode* next = pos->next;
    pos->data = x;
    node->next = next;
    pos->next = node;
}
//约瑟夫环,数到x的人就要出局，返回剩下的最后一个人
ListNode* Josephus(ListNode* hus,int x){
    assert(hus);
    while(hus->next!=hus){
        int count = x;
        while(--count){
            hus = hus->next;
        }
        ListNode* next = hus->next;
        hus->data = next->data;
        hus->next = next->next;
        free(next);
        return hus;
    }
}

//反转链表1->2->3->4->5->6->NULL ------ NULL<-1<-2<-3<-4<-5
//思路：定义三个指针，将其分别遍历整个链表并且反转
ListNode* reverseList(ListNode** node){
    assert(ndoe);
    ListNode* n0 = NULL;
    ListNode* n1 = *head;
    ListNode* n2 = n1->next;
    while(n1){
        n1->next = n0;
        n0 = n1;
        n1 = n2;
        if(n2)
          n2 = n2->next;
    }
    //返回头结点
    *node = n0;
}
//合并两个有序的链表

ListNode* mergeTwoList(ListNode* l1,ListNode* l2){
    ListNode* tmp = new ListNode(0);
    ListNode*　list = tmp->next;
    if(l1==NULL&& l2!=NULL)
      return l2;
    if(l1!=NULL&& l2==NULL)
      return l1;
    while(l1&&l2){
        if(l1->data<l2->data){
            list->next = l1;
            l1 = l1->next;
        }else{
            list->next=l2;
            l2 = l2->next;
        }
        list = list->next;
    }
    if(l1)
      list->next = l1;
    if(l2)
      list->next = l2;
    return tmp->next;
}
//查找单链表的中间节点，只能遍历一次------思路：使用快慢指针进行遍历，快指针走的步数是慢指针的2倍
ListNode* findMidNode(ListNode* ndoe){
    ListNode* slow = node;
    ListNode* fast = node;
    while(fast&& fast->next){
        slow=slow->next;
        fast = fast->next->next;
    }
    return slow;
}
//查找单链表中的倒数第k个节点,------思路：先让快指针走k步，再让快慢指针一起走直到快指针走到结尾
ListNode*  findTailKNode(ListNode* node,int k){
    assert(node);
    ListNode* slow = node;
    ListNode* fast = node;
    while(--k){
        if(fast)
          fast=fast->next;
        else
          return NULL;
    }
    while(fast){
        slow = slow->next;
        fast = fast->next;
    }
    return slow;
}
//判断单链表是否带环----思路：使用快慢指针进行遍历，若快慢指针相遇则说明是带环的
bool isCycle(ListNode* node){
    ListNode* slow = node;
    ListNode* fast = node;
    while(fast&& fast->next){
        slow = slow->next;
        fast = fast->next->next;
        if(slow==fast)
          return true;
    }
    return false;
}
//若链表带环，求环的入口点,思路：将上一个求得带环的相遇的那个节点返回，然后从第一个节点开始进行遍历，相遇的节点和head同时走
//如果相遇了就说明是环的入口。
 ListNode* getEntry(ListNode* head,ListNode* meetNode){
            assert(head);
            if(head!=meetNode){
                head = head->next;
                meetNode = meetNode->next;
                if(head==meetNode)
                  return meetNode;
            }
            return NULL;
 }
//判断两个链表是否相交，如果相交求交点
//思路：先分别遍历两个链表，记录长度，用长链表减去短链表，再让长链表先走差值步，再让短链表一起走，若相遇说明有交点
ListNode* isCross(ListNode* list1,ListNode* list2){
    int count1 = 0, count2 = 0;
    ListNode* l1 = list1;
    ListNode* l2 = list2;
    assert(l1&&l2);
    while(l1){
        count1++;
        l1 = l1->next;
    }
    while(l2){
        count2++;
        l2 = l2->next;
    }
    int k = abs(count1-count2);
    if(count1>count2){
        while(--k)
          list1 = list1->next;
    }
    if(count2>count1){
        while(--k)
          lsit2 = list2->next;
    }
    while(list1&&list2){
        if(list1==list2){
            return list1;
        }
        list1 = list1->next;
        list2 = list2->next;
    }
    return NULL;
}
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