Esempio n. 1
0
void testReverse(void){

	printf("Testing reverse!\n");

	//reverse the list once...
	List l = setUpList();
	printList(l);
	reverseList(l);
	printList(l);


	assert(getElement(l, 0) == 4);
	assert(getElement(l, 1) == 7);
	assert(getElement(l, 2) == 8);
	assert(getElement(l, 3) == 1);
	assert(listLength(l) == START_LENGTH);

	//reverse it again!
	reverseList(l);
	printList(l);
	assert(getElement(l, 0) == 1);
	assert(getElement(l, 1) == 8);
	assert(getElement(l, 2) == 7);
	assert(getElement(l, 3) == 4);
	assert(listLength(l) == START_LENGTH);

	//testing that appended items are reversed as well
	append(l, 32);
	reverseList(l);
	assert(getElement(l, 0) == 32);
	assert(listLength(l) == START_LENGTH + 1);

	printf("Reversing passed!\n");
}
void testReverseList() {
    cout << endl;
    cout << "Test reverseList():" << endl;
    cout << "===================" << endl;

    char a[] = "a";
    char b[] = "b";
    char c[] = "c";
    char d[] = "d";
    char e[] = "e";
    char f[] = "f";
    char g[] = "g";

    Node<char*> head;
    head.data = a;
    head.next = nullptr;
    appendToList(&head, &b[0]);
    appendToList(&head, &c[0]);
    appendToList(&head, &d[0]);
    appendToList(&head, &e[0]);
    appendToList(&head, &f[0]);
    appendToList(&head, &g[0]);
    printList(&head);

    Node<char*>* rev = reverseList(&head);
    printList(rev);

    Node<char*>* orig = reverseList(rev);
    printList(orig);
}
int main(int argc, char **argv)
{
    int i;
    int n;
    NodePtr node;
    JobPtr job;
    ListPtr list;

    if (argc != 2) {
        fprintf(stderr, "Usage: %s <list size> \n",argv[0]);
        exit(1);
    }
    n = atoi(argv[1]);

    list = createList();
    for (i=0; i<n; i++) {
        job = createJob(i, "cmd args");
        node = createNode(job);
        addAtFront(list, node);
    }

    printList(list);
    reverseList(list);
    printList(list);
    reverseList(list);
    printList(list);

    exit(0);
}
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        if (head == NULL)
            return NULL;

        if (head->next == NULL || m == n)
            return head;

        m--; n--;

        if (m == 0)
        {
            head = reverseList(head, n - m + 1);
            return head;
        }

        int c = 0;
        ListNode* curNode = head;
        ListNode* tail = NULL;
        while (c < m)
        {
            assert(curNode);
            tail = curNode;
            curNode = curNode->next;
            c++;            
        }

        curNode = reverseList(curNode, n - m + 1);
        tail->next = curNode;

        return head;
    }
Esempio n. 5
0
    // O(n)-time, O(1)-space
    bool isPalindrome2(ListNode *head) {
        if (head == NULL) {
            return true;
        }

        // 计算中间节点
        ListNode *mid = findMiddleNode(head);

        // 反转后半段链表
        ListNode *reverseHead = reverseList(mid->next);

        // 连接前后两半段链表
        mid->next = reverseHead;

        // 比较前后两个半段链表
        // 注意:后半段链表节点个数 <= 前半段链表节点个数
        bool isPal = true;
        while (reverseHead != NULL) {
            if (head->val != reverseHead->val) {
                isPal = false;
            }
            head = head->next;
            reverseHead = reverseHead->next;
        }

        // 恢复原链表
        reverseHead = reverseList(mid->next);
        mid->next = reverseHead;

        return isPal;
    }
Esempio n. 6
0
/* Deletes nodes which have a node with greater value node on left side */
void delLesserNodes(struct Node **head_ref)
{
	/* 1) Reverse the linked list */
	reverseList(head_ref);
  /*  2) In the reversed list, delete nodes which have a node with greater value node on left side. 
      Note that head node is never deleted because it is the leftmost node.*/
	_delLesserNodes(*head_ref);
  /* 3) Reverse the linked list again to retain the original order */
	reverseList(head_ref);
}
Esempio n. 7
0
File: main.cpp Progetto: CCJY/coliru
void linkedListTest() {
    node * list = nullptr;
    print(list);
    std::cout << "↔";
    reverseList(list);
    print(list);
    for (const auto & elem : {5,1,3,2,4,8,6,9,7}) {
        appendList(list, elem);
        print(list);
        std::cout << "↔";
        reverseList(list);
        print(list);
    }
}
 bool isPalindrome(ListNode* head) {
     if (head == nullptr || head->next == nullptr) return true;
     ListNode* slow = head;
     ListNode* fast = head;
     while (fast->next != nullptr && fast->next->next != nullptr) {
         fast = fast->next->next;
         slow = slow->next;
     }
     ListNode* right = reverseList(slow->next);
     slow->next = nullptr;
     bool result = compare(head, right);
     slow->next = reverseList(right);
     return result;
 }
Esempio n. 9
0
void test( int* arr, int size ) {
  node* root = newList( arr, size );
  print( root );
  reverseList( &root );
  print( root );
  freeList( root );
}
Esempio n. 10
0
    void reorderList(ListNode* head) {
        if (head==NULL || head->next==NULL || head->next->next==NULL)
        {
            return ;
        }
        
        ListNode* l1=head;
        ListNode* l2=getMid(head);
        ListNode* reverseL2=reverseList(l2);
        
		ListNode* newhead=new ListNode(-1);
        ListNode* tail=newhead;
        while (l1!=NULL && reverseL2!=NULL)
        {
			tail->next=l1;
            l1=l1->next;
			tail=tail->next;
            tail->next=reverseL2;
			tail=tail->next;
            reverseL2=reverseL2->next;
        }
       if (l1!=NULL)
       {
		   tail->next=l1;
       }
	   if (reverseL2!=NULL)
	   {
		   tail->next=reverseL2;
	   }
	   head=newhead->next;
    }
Esempio n. 11
0
int main() {
    struct node * head   = NULL;
    struct node * second = NULL;
    struct node * third  = NULL;

    head   = (struct node *) malloc(sizeof(struct node));
    second = (struct node *) malloc(sizeof(struct node));
    third  = (struct node *) malloc(sizeof(struct node));

    head->data = 1;
    head->next = second;

    second->data = 2;
    second->next = third;

    third->data = 3;
    third->next = NULL;

    printf("Normal list order:\n");
    printList(head);
    printf("\nReversed order:\n");
    reverseList(&head);
    printList(head);

    return 0;
}
Esempio n. 12
0
 ListNode* reverseList(ListNode* head) {
     if (!head || !(head->next)) return head;
     ListNode* tmp = reverseList(head->next);
     head->next->next = head;
     head->next = NULL;
     return tmp;
 }
Esempio n. 13
0
 ListNode* reverseList(ListNode *head, int k) {
     ListNode *prev, *current, *next, *tail;
     current = head, prev = NULL, tail = head;
     int cnt = 0;
     while (current != NULL) {
         if (cnt > k)
             break;
         current = current->next;
         cnt++;
     }
     if (cnt < k)
         return head;
     cnt = 0;
     current = head;
     while (cnt < k && current != NULL) {
         next = current->next;
         current->next = prev;
         prev = current;
         current = next;
         cnt++;
     }
     if (current != NULL)
         tail->next = reverseList(current, k);
     return prev;
     
 }
Esempio n. 14
0
int main()
{
    node *listHead = create();
    if (!listHead)
        return 0;
    for (int i = 0; i < 10; i++)
        add(listHead, i);
    deleteNode(listHead, 5);
    insertNode(listHead, 5, 50);

    printf("Length = %d\n", length(listHead));
    printList(listHead);
    reverseList(listHead);
    printList(listHead);

    node *pMid;
    pMid = getMid(listHead);
    printf("Mid data is %d\n", pMid->data);
    add(listHead, 99);
    printList(listHead);
    pMid = getMid(listHead);
    printf("Mid data is %d\n", pMid->data);

    destroy(listHead);
    return 0;
}
Esempio n. 15
0
void reorderList(ListNode* head) 
{
	if (head == NULL || head->next == NULL) return;

	ListNode * slow = head;
	ListNode * fast = head->next;

	while (fast && fast->next)
	{
		fast = fast->next->next;
		slow = slow->next;
	}

	ListNode * half = slow->next;
	slow->next = NULL;
	fast = reverseList(half);
	slow = head;
	while (fast)
	{
		ListNode * slowNext = slow->next;
		ListNode * fastNext = fast->next;
		slow->next = fast;
		fast->next = slowNext;
		slow = slowNext;
		fast = fastNext;
	}
}
Esempio n. 16
0
 ListNode* reverseList(ListNode* head){
     if(!head || !head->next) return head;
     ListNode* newHead = reverseList(head->next);
     head->next->next = head;
     head->next = NULL;
     return newHead ;
 }
 ListNode *reverseKGroup(ListNode *head, int k) {
     /* https://oj.leetcode.com/problems/reverse-nodes-in-k-group/
     e.g.,
     Given this linked list: 1->2->3->4->5
     For k = 2, you should return: 2->1->4->3->5
     For k = 3, you should return: 3->2->1->4->5
     */
     
     if (head == NULL || head->next == NULL || k < 2) return head;
     
     ListNode *dummy = new ListNode(0), *tail = dummy;
     
     while (head != NULL) {
         ListNode *tmp = head;
         for (int i = 1; i < k && tmp != NULL; i++) {
             tmp = tmp->next;
         }
         if (tmp == NULL) {
             tail->next = head;
             break;
         }
         
         ListNode *tmp2 = tmp->next;
         tmp->next = NULL;
         reverseList(head);
         tail->next = tmp;
         tail = head;
         head = tmp2;
     }
     
     head = dummy->next;
     delete dummy;
     return head;
 }
ListNode* reverseList(ListNode* head) {
    if (head == nullptr || head->next == nullptr) return head;
    ListNode* node = reverseList(head->next);
    head->next->next = head;
    head->next = nullptr;
    return node;
}
Esempio n. 19
0
int main(void)
{
    int i = 0, ret = 0;
    struct node *ptr = NULL;

    print_list();

    for(i = 0; i<100000000; i++)
        add_to_list(i);
    
	//print_list();

	ptr = search_in_list(99999999, NULL);
	if(NULL == ptr)
	{
    	printf("Search [val = %d] failed, no such element found\n",i);
    }
    else
    {
        printf("Search passed [val = %d]\n",ptr->val);
    }

    printf("Lets reverse the linked list\n");

    reverseList();

    return 0;
}
 ListNode *reverseBetween(ListNode *head, int m, int n) {
     ListNode *list = head;
     if (m != n) {
         ListNode *pm = NULL, *prem = NULL, *pn = NULL, *nextn = NULL;
         int index = 1;
         ListNode *p = head, *pre = NULL;
         // 找到m、n两个结点
         while (index <= n && p != NULL) {
             if (index == m) {
                 prem = pre;
                 pm = p;
             }
             if (index == n) {
                 pn = p;
                 nextn = p->next;
             }
             index++;
             pre = p;
             p = p->next;
         }
         // 倒序链表 m --> ... -> n
         pn->next = NULL;
         ListNode *from, *to;
         reverseList(pm, &from, &to);
         // 加上头尾
         to->next = nextn;
         if (prem == NULL) {
             list = from;
         } else {
             prem->next = from;
         }
     }
     return list;
 }
Esempio n. 21
0
int main()
{
    struct ListNode *p1,*p2,*head;
    for(int i=0;i<2;i++)
    {
        if(i ==0)
        {
            p1=(struct ListNode *)malloc(sizeof(struct ListNode));
            p1->val=i;
            p1->next=NULL;
            head=p1;
        }
        else
        {
            p2=(struct ListNode *)malloc(sizeof(struct ListNode));
            p2->val=i;
            p2->next=NULL;
            p1->next=p2;
            p1=p1->next;
        }
    }
    printf("origin:");
    printlist(head);
    p2=reverseList(head);
    printf("after:");
    printlist(p2);
}
Esempio n. 22
0
int main(int argc, char **argv)
{
	int i;
	int n;
	struct node *list;
	struct node *node;

	if (argc != 2) {
		fprintf(stderr, "Usage: %s <list size> \n",argv[0]);
		exit(1);
	}
	n = atoi(argv[1]);

	list = NULL;
	for (i=0; i<n; i++) {
		node = (struct node*) malloc(sizeof(struct node));
		if (node == NULL) {
			printf("Error allocating node for linked list\n");
			exit(1);
		}
		node->item = i;
		list = addAtFront(list, node);
	}

	printList(list);

	list=reverseList(list);
	printList(list);

	return 0;
}
 void reorderList(ListNode *head) {
     //step 1: count 
     //iterator headPtr = head;
     int len = lengthOfList( head );
     if( len < 3 ){
         return;
     }
     
     //step 2: reverse
     int mid = (len+1)/2;
     iterator midPtr = getMid( head, mid );  
     iterator mn = midPtr->next;
     midPtr->next = NULL;
     midPtr = reverseList( mn );
     
     //step 3: merge
     iterator hn = head->next;
     mn = midPtr->next;
     while( mn != NULL ){
         head->next = midPtr;
         midPtr->next = hn;
         head = hn;
         midPtr = mn;
         hn = head->next;
         mn = midPtr->next;
     }
     head->next = midPtr;
     midPtr->next = hn;
         
     return ;
 }
Esempio n. 24
0
    bool isPalindrome(ListNode* head) {
        ListNode* slow = head;
        ListNode* fast = head;
        while(fast && fast->next){
            slow = slow->next;
            fast = fast->next->next;
        }
        if(fast) {
            slow = slow->next;
        }

        fast = head;
        slow = reverseList(slow);

        // At this point, the original list is split into two.
        // fast: the 1st half in original order.
        // slow: the 2nd half in reverse order.
        while(slow) {
            if(fast->val != slow->val)
                return false;
            fast = fast->next;
            slow = slow->next;
        }
        return true;
    }
Esempio n. 25
0
 // Iteratively
 // ListNode* reverseList(ListNode* head) {
 //     ListNode* cur = NULL;
 //     for(auto p = head; p != NULL;) {
 //         auto tmp = p->next;
 //         p->next = cur;
 //         cur = p;
 //         p = tmp;
 //     }
 //     return cur;
 // }
 // Recursively
 ListNode* reverseList(ListNode* head) {
     if(!head || !head->next) return head;
     auto p = reverseList(head->next);
     head->next->next = head;
     head->next = NULL;
     return p;
 }
Esempio n. 26
0
bool isPalindrome(ListNode* head) 
{
  if (! head)
    return true;

  int len = ::lenList(head);
  if (len == 1)
    return true;

  ListNode* mid1 = head;
  for (int i=0; i<len/2-1; ++i)
    mid1 = mid1->next;
  ListNode* mid2 = mid1->next;

  ListNode* left = head;
  ListNode* right = mid2;
  mid1->next = NULL;
  if (len % 2)
  {
    right = right->next;
    mid2->next = NULL;
  }
  showList(left);
  showList(right);
  right = reverseList(right);
  showList(right);
  while (left && right)
  {
    if (left->val != right->val)
      return false;
    left = left->next;
    right = right->next;
  }
  return true;
}
    bool isPalindrome(ListNode* head) {
        if (head == NULL)           //empty
            return true;    
        if (head->next == NULL)     //only a head
            return true;

        ListNode* fast = head;      // slow and fast make slow point to the mid
        ListNode* slow = head;
        ListNode* slowPrev = NULL;      // slowPrev is the element before slow  
        while (fast!=NULL && fast->next!=NULL){ // 'fast!=NULL' means the size of the list is odd,'fast->next' means even
            fast = fast->next->next;
            if (fast==NULL || fast->next==NULL){
                slowPrev = slow;      
            }
            slow = slow->next;
        }
        slowPrev->next = NULL;                  //  slowPrev->next points to NULL
        ListNode* tail = reverseList(slow); // reverse the list after the slow pointer(the mid of the list)
        while(head!=NULL && tail!=NULL){
            if (head->val != tail->val){
                delete tail;
                return false;
            }
            head = head->next;
            tail = tail->next;
        }
        delete fast,slow,slowPrev,tail;
        return true;
    }
Esempio n. 28
0
void main(int argc, char** argv){
  node *head = NULL;
  node *test = NULL;
  node *merged = NULL;

  addNode(5, &head);
  addNode(6, &head);
  addNode(7, &head);
  addNode(10, &head);
  
  addNode(4, &test);
  addNode(8, &test);
  addNode(11, &test);

  merged = merge_sorted(head, test);

  printList(merged);

  //head->next->next->next->next = head->next;
 if(detectLoop(head)){
    printf("loop found\n");
    exit(0);
  }
 // printList(head);

  deleteNode(7, head);

 // printList(head);

  head = reverseList(head, NULL);

  //printList(head);

  destroyList(head);
}
Esempio n. 29
0
File: 2.1.c Progetto: kun-g/sketch
PNODE subRevList(PNODE head, int revCount) {
  PNODE p = head;
  int count = 0;
  PNODE curHead = p;
  PNODE prevTail = NULL;

  while (p) {
    count += 1;
    if (count%revCount == 0) {
      if (prevTail) {
        prevTail->next = p;
        prevTail->next_address = p->address;
      } else {
        head = p;
      }

      PNODE nextHead = p->next;
      p->next = NULL;
      p->next_address = -1;
      reverseList(curHead);
      prevTail = curHead;
      curHead = nextHead;
      p = nextHead;
    } else {
      p = p->next;
    }
  }
  if (count%revCount && prevTail) {
    prevTail->next = curHead;
    prevTail->next_address = curHead->address;
  }

  return head;
}
Esempio n. 30
0
int main(int argc, char** argv){
    struct ListNode *head, *probe;
    int lstSize[4] = {0,1,2,8};
    int i;
    srand(time(NULL));
    for (i = 0; i < 4; i++){
        int j;
        /* Initialize the linked list. */
        head = NULL;
        head = (struct ListNode *)malloc(sizeof(struct ListNode));
        head->val = 0x7a636664;
        head->next = NULL;
        for (probe = head, j = 0; j < lstSize[i]; j++){
            probe->next = (struct ListNode *)malloc(sizeof(struct ListNode));
            probe = probe->next;
            probe->val = rand() % 100;
            probe->next = NULL;
        }
        printf("Case%2d: \n",i+1);
        printf("The original linked list:\n");
        printLst(head);
        /* Test */
        printf("The reversed linked list:\n");
        reverseList(head);
        printLst(head);
        printf("\n");
        deleteLst(head);
    }
	system("pause");
    return 0;
}