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Cover image for Day 3-4| Linked List: leetcode 203,706,206,24,19,02.07,142 in Cpp
Jiayi Ouyang
Jiayi Ouyang

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Day 3-4| Linked List: leetcode 203,706,206,24,19,02.07,142 in Cpp

#algorithms #cpp

Linked List is a linear structure linked by pointers.

It consists of nodes where each node contains data and a reference (link) to the next node in the sequence.

This allows for dynamic memory allocation and efficient insertion and deletion operations compared to arrays.
The entry node of the list is head

Linked List Vs Arrays:

Array Linked List
Data Structure Contiguous Non-contiguous
Memory Allocation Static Dynamic
Insertion/Deletion Inefficient O(n) Efficient O(1)
Access Random O(1) Sequential O(n)
Scenario data is constant, frequently access, rarely add/remove data varies over time, frequently add/remove, rarely access

Types of Linked List:

  1. Singly Linked List
  2. Doubly Linked List
  3. Circular Linked List

General Class Definition 

struct ListNode
{
    int val; // data
    ListNode * next;
    ListNode(int x) : val(x), next(nullptr) {}
}
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dummy head node

It's very helpful to declare a dummy head whose next points the head of original list.
In this way, we can handle the situation when deleting the original head, the dummyhead->next is always the actual head.
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203 Remove Linked List Elements

Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, return the new head

ListNode* removeElements(ListNode* head, int val) {
    ListNode* dummyHead = new ListNode(0, head);
    ListNode* cur = dummyHead;
    while(cur->next != nullptr)
    {
        if(cur->next->val == val)
        {
            ListNode* toRemove = cur->next;
            cur->next = cur->next->next;
            delete toRemove;
        }
        else
            cur = cur->next;
    }
    return dummyHead->next;
}
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707 Design Linked List

Design your implementation of the linked list. Assume all nodes in the linked list are 0-indexed.

Implement the MyLinkedList class:

  • MyLinkedList() Initializes the MyLinkedList object.
  • int get(int index) Get the value of the indexth node in the linked list. If the index is invalid, return -1.
  • void addAtHead(int val) Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
  • void addAtTail(int val) Append a node of value val as the last element of the linked list.
  • void addAtIndex(int index, int val) Add a node of value val before the indexth node in the linked list. If index equals the length of the linked list, the node will be appended to the end of the linked list. If index is greater than the length, the node will not be inserted.
  • void deleteAtIndex(int index) Delete the indexth node in the linked list, if the index is valid.

Singly Linked List

Using dummy head as a data member, please pay attention to initialization of cur:

  1. when addAtIndex(index), we want to do cur->next = new .., suppose index = 0, the cur has to be dummyHead, so ListNode* cur = dummy
  2. when get(index), we wanna return cur->value;, when index = 0, the cur has to be at true head, so ListNode* cur = dummyHead->next; 
class MyLinkedList {
public:

    struct ListNode
    {
        int val;
        ListNode* next;
        ListNode(int x) : val(x), next(nullptr) {}
        ListNode(int x, ListNode* nextNode) : val(x), next(nextNode) {}
    };

    MyLinkedList() : size(0) {
        dummyHead = new ListNode(0);
    }

    // index 0 refers to (dummyHead -> next)
    // so 0 <--> dummyHead->next
    int get(int index) {

        if(index > size - 1 || index < 0)    return -1;

        ListNode* cur =  dummyHead->next;
        while(index--)
        {
            cur = cur -> next;
        }
        return cur ? cur->val : -1;
    }

    void addAtHead(int val) {

        dummyHead->next = new ListNode(val, dummyHead->next);
        ++size;
    }

    void addAtTail(int val) {
        ListNode * cur = dummyHead;
        while(cur->next)
        {
            cur = cur->next;
        }
        cur->next = new ListNode(val);
        ++size;
    }

    // add at 0, cur has to be dummyHead to be able to assign the head
    // so 0 <--> dummyHead
    void addAtIndex(int index, int val) {
        if(index > size)    return;

        ListNode * cur = dummyHead;
        while(index--)
        {
            cur = cur->next;
        }
        ListNode* next = cur->next;
        cur->next = new ListNode(val, next);
        ++size;
    }

    // to delete at 0, cur has to be at dummyHead
    // so 0 <--> dummyHead
    void deleteAtIndex(int index) {

        if(index > size - 1 || index < 0)
            return;

        ListNode* cur = dummyHead;
        while(index--)
        {
            cur = cur -> next;
        }
        ListNode* next = cur->next;
        cur->next = next->next;
        --size;
        delete next;
    }

private:
    ListNode* dummyHead;
    int size;
};
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Doubly Linked List 

class MyLinkedList {
public:

    struct ListNode
    {
        int val;
        ListNode* next;
        ListNode* pre;
        ListNode(int x) : val(x), next(nullptr), pre(nullptr) {}
        ListNode(int x, ListNode* nextNode, ListNode* preNode) : val(x), next(nextNode), pre(preNode) {}
    };

    MyLinkedList() : size(0) {
        dummyHead = new ListNode(0);
    }

    // index 0 refers to (dummyHead -> next)
    // so 0 <--> dummyHead->next
    int get(int index) {
        if(index > size - 1 || index < 0)    return -1;
        ListNode* cur =  dummyHead->next;
        while(index--)
        {
            cur = cur -> next;
        }
        return cur ? cur->val : -1;
    }

    void addAtHead(int val) {
        ListNode* next = dummyHead->next;
        dummyHead->next = new ListNode(val, next, dummyHead);
        if(next)
            next->pre = dummyHead->next;
        ++size;
    }

    void addAtTail(int val) {
        ListNode * cur = dummyHead;
        while(cur->next)
        {
            cur = cur->next;
        }
        cur->next = new ListNode(val);
        cur->next->pre = cur;
        ++size;
    }

    // add at 0, cur has to be dummyHead to be able to assign the head
    // so 0 <--> dummyHead
    void addAtIndex(int index, int val) {
        if(index > size)    return;
        ListNode * cur = dummyHead;
        while(index--)
        {
            cur = cur->next;
        }
        ListNode* next = cur->next;
        cur->next = new ListNode(val, next, cur);
        if(next)
            next->pre = cur->next;
        ++size;
    }

    // to delete at 0, cur has to be at dummyHead
    // so 0 <--> dummyHead
    void deleteAtIndex(int index) {
        if(index > size - 1 || index < 0)
            return;
        ListNode* cur = dummyHead;
        while(index--)
        {
            cur = cur -> next;
        }
        ListNode* next = cur->next;
        cur->next = next->next;
        if(next->next)
        {
            next->next->pre = cur;
        }
        --size;
        delete next;
    }
private:
    ListNode* dummyHead;
    int size;
};
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206 Reverse Linked List

Given the head of a singly linked list, reverse the list, and return the reversed list.

Method 1 dummy head

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 ListNode* reverseList(ListNode* head) {
    ListNode* dummyHead = new ListNode(0);
    ListNode* cur = head;
    while(cur)
    {
        ListNode* newHead = cur;
        ListNode* newHeadNext = dummyHead->next;
        cur = cur->next;
        dummyHead->next = newHead;
        newHead->next = newHeadNext;
    }
    return dummyHead->next;
}
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Method 2 Pre and Cur Pointer

Quite same as method 1
Reversing the linked list with new head moving from original head to tail

ListNode* reverseList(ListNode* head) {
    ListNode * pre = nullptr;
    ListNode * cur = head;
    ListNode * nextIter =  nullptr;
    while(cur)
    {
        nextIter = cur->next;
        cur->next = pre;
        pre = cur;
        cur = nextIter;
    }
    return pre;
}
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Method 3 recursion with same idea from method 2 

ListNode* reverse(ListNode* pre,ListNode* cur){
    if(cur == NULL) return pre;
    ListNode* nextIter = cur->next;
    cur->next = pre;
    // same as method 2:
    // pre = cur;
    // cur = temp;
    return reverse(cur,temp);
}

ListNode* reverseList(ListNode* head)
{
    // same as method 2:
    // pre = nullptr
    // cur = head
    return reverse(nullptr, head);
}
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Method 4 recursion from tail to head

  1. Keeping looking for the tail which is the new head, returns for the first time,
  2. head->next->next = nullptr because it's the new tail
  3. Reversing the head and head->next

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ListNode* reverseList(ListNode* head) {
    // finding tail which is new head
    if(head == NULL) return NULL;
    if (head->next == NULL) return head;
    ListNode *newHead = reverseList(head->next);

    // imagine you have the newhead, reverse the last second
    head->next->next = head;
    // 此时的 head 节点为尾节点，next 需要指向 NULL
    head->next = NULL;
    return newHead;
}
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Simulation of the recursion

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24 Swap Nodes in Pairs

Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)

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ListNode* swapPairs(ListNode* head) {

    if(!head || !head->next)    return head;
    ListNode* dummyHead = new ListNode(0, head);

    ListNode * cur = dummyHead;
    ListNode* Swap1Node;
    ListNode* NextPair;

    while(cur->next && cur->next->next)
    {
        Swap1Node = cur->next;
        NextPair = cur->next->next->next;
        cur->next = cur->next->next; // Swap2Node
        cur->next->next = Swap1Node;
        cur->next->next->next = NextPair;
        cur = Swap1Node;
    }
    return dummyHead->next;

}
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Dual Pointers Method in Linked List

19 Remove Nth Node From End of List -- fast and slow

Given the head of a linked list, remove the nth node from the end of the list and return its head.

ListNode* removeNthFromEnd(ListNode* head, int n) {

    ListNode* dummyHead = new ListNode(0, head);
    ListNode* slow = dummyHead;
    ListNode* fast = dummyHead;
    while(n-- > -1 && fast) // fast move one more, 
    {                       //since slow has to be front of the last-nth
        fast = fast->next;
    }

    while(fast) // when fast gets to the tail, slow is at the front of
    {           // the last n-th node
        fast = fast->next;
        slow = slow->next;
    }

    ListNode* tmp = slow->next;
    slow->next = tmp->next;
    delete tmp;

    return dummyHead->next;

}
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02.07 Intersection of Two Linked Lists LCCI

Given two (singly) linked lists, determine if the two lists intersect. Return the inter­secting node. Note that the intersection is defined based on reference, not value. That is, if the kth node of the first linked list is the exact same node (by reference) as the jth node of the second linked list, then they are intersecting.

ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {

    // 1. get lengths of two lists
    int lenA = 0, lenB = 0;
    {
        ListNode* curA = headA;
        ListNode* curB = headB;
        while(curA)
        {
            ++lenA;
            curA = curA->next;
        }

        while(curB)
        {
            ++lenB;
            curB = curB->next;
        }
    }

    // 2. distinguish which list is longer
    ListNode* longer, * shorter;
    int diff = 0;
    if(lenA < lenB)
    {
        diff = lenB - lenA;
        longer = headB;
        shorter = headA;
    }
    else
    {
        diff = lenA - lenB;
        longer = headA;
        shorter = headB;
    }

    // 3. move longer pointer in the longer list, to make longer and shorter 
    // pointers at the same distance to tail
    while(diff-- && longer)
    {
        longer = longer->next;
    }

    // 4. move dual pointers together until they meet
    while(longer && shorter && longer != shorter)
    {
        longer = longer->next;
        shorter = shorter->next;
    }

    // 5. there might be no intersetion at all, so be careful when returning
    return longer == shorter ? longer : NULL;
}
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142 Linked List Cycle II

Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.

Do not modify the linked list.

Image description

2 * (x + y) = x + n * (y + z) + y
x + y = n * (y + z)
x = (n - 1) * (y + z) + z // (y + z) is the length of circle.
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ListNode *detectCycle(ListNode *head) {
    ListNode* fast = head, *slow = head;
    while(fast && fast->next)
    {
        fast = fast->next->next;
        slow = slow->next;
        if(fast == slow) // get to meeting point
        {
            break;
        }
    }

    if(!fast || !fast->next)    return NULL;

    fast = head;
    while(fast != slow) // x = z, so fast and slow will eventually meet at 
    {                   // the start of intersection
        fast = fast->next;
        slow = slow->next;
    }
    return fast;
}
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