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算法原理与实践（链表）

分类: IT文章 • 2022-05-03 11:11:37

　　链表

　　大纲
　　1. 链表介绍
　　2. 基本操作
　　3. Dummy Node
　　4. 追赶指针技巧
　　5. 例题分析

　　链表介绍
　　单向链表(singly linked list)，每个节点有一个 next 指针指向后一个节点，还有一个成员变量用以储存数值；
　　双向链表(Doubly Linked List)，还有一个 prev 指针指向前一个节点。
　　Search: O(n), Del, Add: O(1)

　　Remove Duplicates from Sorted List
　　Given a sorted linked list, delete all duplicates such that each element appear only once. For example, Given 1->1->2, return 1->2. Given 1->1->2->3->3, return 1->2->3.

#include <list>
#include <iostream>
#include <iterator>

using namespace std;

template<typename T>
void removeDuplicate(list<T>& l)
{
    list<T>::iterator chaser = l.begin();
    list<T>::iterator runner = ++l.begin();

    while (runner != l.end())
    {
        if (*chaser == *runner)
        {
            runner = l.erase(runner);
        }
        else
        {
            ++chaser;
            ++runner;
        }
    }
}

template<typename T>
void dump(list<T> l)
{
    for (auto &e : l)
    {
        cout << e << " ";
    }
    cout << endl;
}

int main()
{
    list<int> l = { 1, 1, 2 };
    removeDuplicate(l);
    dump(l);

    list<int> l2 = { 1, 1, 2, 3, 3 };
    removeDuplicate(l2);
    dump(l2);
}
View Code

　　Remove Duplicates from Sorted List II
　　Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
　　For example,
　　Given 1->2->3->3->4->4->5, return 1->2->5.
　　Given 1->1->1->2->3, return 2->3.

#include <iostream>
#include <list>
#include <unordered_map>

using namespace std;

template<typename T>
void removeDuplicate2(list<T>& l)
{
    unordered_map<T, unsigned> m;

    for (auto &e : l)
        m[e]++;

    list<T>::iterator it = l.begin();

    while (it != l.end())
    {
        if (m[*it] > 1)
            it = l.erase(it);
        else
            it++;
    }
}

template<typename T>
void dump(const list<T>& l)
{
    for (auto &e : l)
    {
        cout << e << " ";
    }
    cout << endl;
}

int main()
{
    list<int> l = { 1, 2, 3, 3, 4, 4, 5 };

    removeDuplicate2(l);

    dump(l);


    list<int> l2 = { 1, 1, 1, 2, 3 };

    removeDuplicate2(l2);

    dump(l2);
}
View Code

　　Partition List
　　Given a linked list and a value x, write a function to reorder this list such that all nodes less than x come before the nodes greater than or equal to x.
　　解题分析：将list分成两部分，但两部分的head节点连是不是null都不确定。但总是可以创建两个dummy节点然后在此基础上append，这样就不用处理边界条件了。

#include <iostream>
#include <list>

using namespace std;

template<typename T> 
list<T>* partitionList(const list<T>& l, const T& val)
{
    list<T> rightList;
    list<T>* leftList = new list<T>;

    for (auto &e : l)
    {
        if (e > val)
            rightList.push_back(e);
        else
            leftList->push_back(e);
    }

    leftList->insert(leftList->end(), rightList.cbegin(), rightList.cend());

    return leftList;
}

template<typename T>
void dump(const list<T>& l)
{
    for (auto &e : l)
        cout << e;

    cout << endl;
}

int main()
{
    list<int> l = { 6, 5, 4, 3, 2, 1 };

    list<int>* pL = partitionList(l, 3);

    dump(*pL);

    return 0;
}
View Code

　　

　　追赶指针技巧

　　对于寻找list某个特定位置的问题，不妨用两个变量 chaser 与 runner，以不同的速度遍历 list，找到目标位置： ListNode *chaser = head, *runner = head。并且可以用一个简单的小 test case 来验证（例如长度为4和5的list）

　　Middle Point

　　Find the middle point of linked list.
　　解题分析： 寻找特定位置，runner以两倍速前进，chaser 一倍速，当runner到达tail时，chaser即为所求解。

#include <list>
#include <iostream>

using namespace std;

template<typename T>
typename list<T>::iterator findMiddlePoint(list<T> &l)
{
    list<T>::iterator itRunner = l.begin();
    list<T>::iterator itChaser = l.begin();

    while (itRunner != l.end())
    {
        itRunner++;

        if (itRunner != l.end())
        {
            itRunner++;
            itChaser++;
        }
    }

    return itChaser;
}

int main()
{
    list<int> l = { 1, 2, 3, 4, 5 };

    cout << *findMiddlePoint(l) << endl;

    return 0;
}
View Code

　　kth to Last element
　　Find the kth to last element of a singly linked list
　　解题分析：之前类似。只是runner与chaser以相同倍速前进，但runner提前k步出发

#include <list>
#include <iostream>

using namespace std;

template<typename T>
typename list<T>::iterator findKthLast(list<T> &l,const size_t k)
{
    list<T>::iterator itRunner = l.begin();
    list<T>::iterator itChaser = l.begin();

    for (size_t i = 0; i < k; i++)
        itRunner++;

    while (itRunner != l.end())
    {
        itRunner++;
        itChaser++;
    }

    return itChaser;
}

int main()
{
    list<int> l = { 1, 2, 3, 4, 5 };

    cout << *findKthLast(l, 1) << endl;

    return 0;
}
View Code

　　

　　如何判断一个单链表中有环？
　　Given a linked list, determine if it has a cycle in it.

算法原理与实践（链表）

/*返回nullptr说明没有环，否则返回指向环节点的指针*/
struct Node* findLoop(struct Node* head) {
    struct Node* runner = head;
    struct Node* chaser = head;

    while (runner != nullptr) {
        runner = runner->next;

        if (runner->next != nullptr)
            runner = runner->next;
        else
            return nullptr;

        if (runner->next == chaser)
            return chaser;
        else
            chaser = chaser->next;
    }

    return nullptr;
}
View Code

　　Circular List Node
　　Given a circular linked list, return the node at the beginning of the loop
　　解题分析：寻找某个特定位置，用 runner technique。Runner 以两倍速度遍历，假定有 loop，那么 runner 与 chaser 一定能在某点相遇。相遇后，再让 chaser 从 head 出发再次追赶 runner，第二次相遇的节点为 loop 开始的位置。（我对 chaser 第二次再从 head 出发追赶这里不赞同，因为第一次就已经找到了环节点，而且第二次再出发时，找到的还是环节点）。

　　代码是只相遇一次，并且相遇的节点就是 loop node。

#include <iostream>

using namespace std;

const unsigned SIZE = 5;
typedef int DateType;

struct Node {
    DateType val;
    struct Node *next;
};

void initList(struct Node **head) {
    struct Node* curr = (*head);

    for (int i = 0; i < SIZE; i++) {
        curr->next = new struct Node();
        curr = curr->next;
        curr->val = i;
        curr->next = nullptr;
    }
}

void print(struct Node* head) {
    unsigned loop = 0;
    struct Node* curr = head;
    while (curr != nullptr) {
        cout << curr->val << " ";
        curr = curr->next;
        if (++loop > SIZE * 2)
            break;
    }
}

/*返回nullptr说明没有环，否则返回指向环节点的指针*/
struct Node* findLoop(struct Node* head) {
    struct Node* runner = head;
    struct Node* chaser = head;

    while (runner != nullptr) {
        runner = runner->next;

        if (runner->next != nullptr)
            runner = runner->next;
        else
            return nullptr;

        if (runner->next == chaser)
            return chaser;
        else
            chaser = chaser->next;
    }

    return nullptr;
}

void relesseSingleLine(struct Node **head) {
    struct Node* curr = *head;
    struct Node* runner = *head;

    while (curr != nullptr) {
        runner = runner->next;
        delete curr;
        curr = runner;
    }
}

void releaseLoopLine(struct Node **head) {
    while ( (*head)->next != *head) {
        struct Node *runner = (*head)->next->next;
        delete (*head)->next;
        (*head)->next = runner;
    }
    (*head)->next = nullptr;
}

void release(struct Node **head) {
    struct Node *loop = findLoop(*head);
    if (loop != nullptr) {
        //has a loop
        releaseLoopLine(&loop);
    }

    relesseSingleLine(head);
}

/*返回第index个位置上的节点指针*/
struct Node* getPtr(struct Node* head, unsigned index) {
    struct Node* curr = head;
    for (unsigned i = 0; i < index; i++)
        curr = curr->next;

    return curr;
}

int main() {
    struct Node* head = new struct Node();
    initList(&head);

    getPtr(head, SIZE)->next = getPtr(head, 3);

    cout << findLoop(head)->val << endl;

    print(head);
    release(&head);

    return 0;
}
View Code

　　判断两个单链表是否有交点？
　　先判断两个链表是否有环，如果一个有环一个没环，肯定不相交；如果两个都没有环，判断两个列表的尾部是否相等；如果两个都有环，判断一个链表上的 Z 点是否在另一个链表上。
　　如何找到第一个相交的节点？
　　求出两个链表的长度 L1, L2（如果有环，则将 Y 点当做尾节点来算），假设 L1 < L2，用两个指针分别从两个链表的头部开始走，长度为 L2 的链表先走 L2 - L1，然后两个一起走，直到二者相遇。

　　Rotate List

　　Given a list, rotate the list to the right by k places, where k is non-negative.

　　e.g. 

　　example k = 4 and list = 10->20->30->40->50->60.
　　change to => 50->60->10->20->30->40

#include <list>
#include <iostream>

using namespace std;

template<typename T>
list<T> rotateList(const list<T> &l, const size_t k)
{
    list<T>::const_iterator itK = l.cbegin();

    for (size_t i = 0; i < k; i++)
        itK++;

    list<T> l2;

    for (list<T>::const_iterator itRight = itK; itRight != l.cend(); itRight++)
        l2.push_back(*itRight);

    for (list<T>::const_iterator itLeft = l.cbegin(); itLeft != itK; itLeft++)
        l2.push_back(*itLeft);

    return l2;
}

int main()
{
    list<int> l1 = { 1, 2, 3, 4, 5 ,6 };

    list<int> l2 = rotateList(l1, 4);

    for (auto &e : l2)
        cout << e << " ";

    return 0;
}
View Code

　　模式识别
　　1.在遍历 Linked list 时，注意每次循环内只处理一个或一对节点。核心的节点只处理当前这一个，否则很容易出现重复处理的问题。

　　Reverse Linked List
　　Reverse the linked list and return the new head.
　　循环遍历linked-list, 每次只处理当前指针的next 变量。
　　非递归 vs 递归

#include <list>
#include <iostream>
#include <algorithm>

using namespace std;

template<typename T>
list<T> reverseList(list<T> &l)
{
    list<T> l2;

    for (auto &e : l)
        l2.push_front(e);
    
    return l2;
}

int main()
{
    list<int> l = { 1, 2, 3, 4, 5, 6, 7, 8 };

    list<int> l2 = reverseList(l);

    for (auto &e : l2)
        cout << e << " ";

    return 0;
}
View Code　　

　　模式识别
　　2. Swap Node 问题
　　交换两个节点，不存在删除的话，两个节点的prev节点的next指针，以及这两个节点的next指针，会受到影响。总是可以
　　　　a. 先交换两个prev节点的next指针的值；
　　　　b. 再交换这两个节点的next指针的值。
　　无论这两个节点的相对位置和绝对位置如何，以上的处理方式总是成立。

　　Swap Adjacent（邻近的） Nodes
　　Given a linked list, swap every two adjacent nodes and return its head.

#include <list>
#include <iostream>
#include <algorithm>

using namespace std;

template<typename T>
void swapAdjancentNode(list<T> &l)
{
    int tick = 2;

    list<T>::iterator itRunner = l.begin();
    list<T>::iterator itChaser = l.begin();

    itRunner++;

    while (itRunner != l.end())
    {
        if (tick == 2)
        {
            swap(*itRunner, *itChaser);
            tick = 0;
        }

        itRunner++;
        itChaser++;
        tick++;
    }
}

int main()
{
    list<int> l = { 1, 2, 3, 4, 5, 6, 7};

    swapAdjancentNode(l);

    for (auto &e : l)
        cout << e << " ";

    return 0;
}
View Code

　　模式识别
　　3. 同时处理两个 linked list 的问题，循环的条件一般可以用 while( l1 && l2 ) ,再处理剩下非 NULL 的 list。这样的话，边界情况特殊处理，常规情况常规处理。 

　　Add List Sum
　　Given two linked lists, each element of the lists is a integer. Write a function to return a new list, which is the “sum” of the given two lists.
　　　　Part a. Given input (7->1->6) + (5->9->2), output 2->1->9.
　　　　Part b. Given input (6->1->7) + (2->9->5), output 9->1->2.
　　解题分析：对于 a，靠前节点的解不依赖靠后节点，因此顺序遍历求解即可。对于 b，靠前节点的解依赖于靠后节点（进位），因此必须用递归或栈处理。并且，subproblem 返回的结果，可以是一个自定义的结构（进位 + sub-list）。当然，也可以 reverse List 之后再用 a 的解法求解。

#include <list>
#include <iostream>

using namespace std;

list<int> addListSum(const list<int> l1, const list<int> l2)
{
    list<int> l;

    int carry = 0;

    list<int>::const_iterator it1 = l1.cbegin();
    list<int>::const_iterator it2 = l2.cbegin();

    while (it1 != l1.cend() || it2 != l2.cend())
    {
        int sum = carry + (it1 != l1.cend() ? *it1 : 0) + (it2 != l2.cend() ? *it2 : 0);

        carry = sum / 10;

        sum = sum % 10;

        l.push_back(sum);

        if (it1 != l1.cend())
            it1++;

        if (it2 != l2.cend())
            it2++;
    }

    if (carry)
        l.push_back(carry);

    return l;
}

int main()
{
    list<int> l1 = { 7, 1, 6 };
    list<int> l2 = { 5, 9, 2, 1 };


    list<int> l = addListSum(l1, l2);

    for (auto e : l)
        cout << e << " ";

    return 0;
}
View Code

　　Merge Two Sorted List
　　Merge two sorted linked lists and return it as a new list.

template<typename T>
list<T>& merge2SortedList(list<T> &l1, const list<T> &l2)
{
    list<T>::const_iterator it1 = l1.cbegin();
    list<T>::const_iterator it2 = l2.cbegin();

    while (it1 != l1.cend()  &&  it2 != l2.cend())
        (*it1 < *it2) ? it1++ : l1.insert(it1, *it2++);

    while(it2 != l2.cend())
        l1.insert(it1, *it2++);

    return l1;
}
View Code

　　

　　Merge K Sorted List

　　1: ListNode *mergeKLists(vector<ListNode *> &lists) {
　　2: 　　if(lists.size() == 0) return NULL;
　　3: 　　ListNode *p = lists[0];
　　4: 　　for(int i =1; i< lists.size(); i++)
　　5: 　　{
　　6: 　　　　p = merge2Lists(p, lists[i]);
　　7: 　　}
　　8: 　　return p;
　　9: }

#include <list>
#include <iostream>

using namespace std;

template<typename T>
list<T>& merge2SortedList(list<T> &l1, const list<T> &l2)
{
    list<T>::const_iterator it1 = l1.cbegin();
    list<T>::const_iterator it2 = l2.cbegin();

    while (it1 != l1.cend()  &&  it2 != l2.cend())
        (*it1 < *it2) ? it1++ : l1.insert(it1, *it2++);

    while(it2 != l2.cend())
        l1.insert(it1, *it2++);

    return l1;
}

template<typename T>
list<T>* mergeMultiSortedList(list<T> *first, list<T> *last)
{
    if (!(last - first))
        return nullptr;

    for (list<T> *it = first + 1; it != last; it++)
        *first = merge2SortedList(*first, *it);

    //*first = merge2SortedList(*first, *last);

    return first;
}

int main()
{
    list<int> l[4];
    l[0] = { 0, 4, 8};
    l[1] = { 1, 5, 9};
    l[2] = { 2, 6, 10, 14 };
    l[3] = { 3, 7, 11, 15, 19 };

    l[0] = *mergeMultiSortedList(l, l + 4);

    for (auto e : l[0])
        cout << e << " ";

    return 0;
}
View Code

　　-Better Solution?
　　HEAP
　　　　1. Create a heap to store ListNode*, which should sort list nodes in the ascending order or node values.
　　　　2. Insert the first node(head) of each list into the heap, so that we store all the k entries in the heap.
　　　　3. Get the top element in heap and add in to the merged list.
　　　　4. If the top element of heap is the last node in a list, pop it and go to step 3.
　　　　5. If the top element of heap has followers, pop it and push its following node into heap.
　　　　6. Go to step 3 until the heap is empty.

　　--代码实现

#include <iostream>
#include <list>
#include <vector>

using namespace std;

template<typename T>
class Heap
{
public:
    Heap() : _v(1) {};

    void insert(T val)
    {
        _v.push_back(val);

        vector<T>::size_type i;

        for (i = _v.size() - 1; _v[i / 2] > val; i /= 2)
        {
            _v[i] = _v[i / 2];
        }

        _v[i] = val;
    }

    T removeMin()
    {
        if (_v.size() == 1)
        {
            return NULL;
        }

        T minElement = _v[1];
        T lastElement = _v[_v.size() - 1];

        vector<T>::size_type i, child;

        for (i = 1; i * 2 < _v.size(); i = child)
        {
            child = i * 2;

            if (child != _v.size() - 1 && _v[child + 1] < _v[child])
            {
                child++;
            }

            if (lastElement > _v[child])
            {
                _v[i] = _v[child];
            }
            else
            {
                break;
            }
        }

        _v[i] = lastElement;
        _v.pop_back();

        return minElement;
    }

    unsigned size()
    {
        return _v.size() - 1;
    }

private:
    vector<T> _v;
};

template<typename T>
list<T> mergeSortList(vector<list<T>> lists)
{
    Heap<T> h;

    for (auto l : lists)
    {
        for (auto e : l)
        {
            h.insert(e);
        }
    }

    list<T> l;

    while (h.size() != 0)
    {
        l.push_back(h.removeMin());
    }

    return l;
}

int main()
{
    vector<list<int>> lists = { {1,5,10}, {2, 6, 11}, {3, 7, 12} };

    for (auto e : mergeSortList<int>(lists))
    {
        cout << e << " ";
    }

    return 0;
}
View Code

　　模式识别
　　4.如果对靠前节点的处理必须在靠后节点之后，即倒序访问问题，则用 recursion（递归），或者等效地，stack来解决。

　　例题

　　Traverse the linked list reversely.

void traverse(ListNode *head) {
    if (head == NULL)
        return;
    traverse(head->next);
    visit(head);
}

　　基础训练
　　1. Insert a Node in Sorted List
　　2. Remove a Node from Linked List
　　3. Reverse a Linked List
　　4. Merge Two Linked Lists
　　5. Find the Middle of a Linked List

　　工具箱
　　C++
　　Doubly linked list 的实现类是 std::list<T>.
　　常用 iterator： begin(), end(), rbegin(), rend().
　　常用函数:empty(), size(), push_back(T value), pop_back(T value); erase(iterator pos), insert(iterator pos, T value);
　　Java
　　Doubly linked list 的实现类是 LinkedList<E>
　　常用函数：add(E e), add(int index, E element), remove(int index), addAll(Collection<? Extends E> c), get(int index),

　　Reorder List
　　Given a singly linked list L: L0→L1→…→Ln-1→Ln, reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→… You must do this in-place without altering the nodes' values.
　　Example
　　For example, Given 1->2->3->4->null, reorder it to 1->4->2->3->null.

#include <iostream>
#include <list>
#include <algorithm>

using namespace std;

template<typename T>
void reorderList(list<T> &l)
{
    list<T>::iterator it1 = ++l.begin();
    list<T>::iterator it2 = --l.end();

    for (size_t i = 0; i < l.size() / 2; i++)
    {
        swap(*it1++, *it2--);
    }
}

int main()
{
    list<int> l = { 0,1,2,3,4,5,6};

    reorderList(l);

    for (auto &e : l)
    {
        cout << e << " ";
    }

    return 0;
}
View Code

　　Clone a linked list with next and random pointer
　　A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null. Return a deep copy of the list.

算法原理与实践（链表）

　　solution 1: using hashmap

算法原理与实践（链表）

　　batter solution ?

算法原理与实践（链表）

　　Homework

　　Remove Duplicates from Unsorted List
　　Write a removeDuplicates() function which takes a list and deletes any duplicate nodes from the list. The list is not sorted.
　　For example
　　if the linked list is 12->11->12->21->41->43->21, then removeDuplicates() should convert the list to 12->11->21->41->43. If temporary buffer is not allowed, how to solve it?

#include <unordered_map>
#include <list>
#include <iostream>

using namespace std;

template<typename T>
list<T> removeDuplicateFromUnsortedList(const list<T> &l)
{
    unordered_map<T, bool> m;
    list<T> l2;

    for (auto &e : l)
    {
        if (m.find(e) == m.end())
        {
            m[e] = true;
            l2.push_back(e);
        }
    }

    return l2;
}

int main()
{
    list<int> l = { 12,11,12,21,41,43,21 };

    list<int> l2 = removeDuplicateFromUnsortedList(l);

    for (auto &e : l2)
        cout << e << " ";

    return 0;
}
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　　Reverse a linked list
　　Reverse a linked list from position m to n.
　　Note
　　Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list.
　　Example
　　Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL.
　　Challenge
　　Reverse it in-place and in one-pass

#include <iostream>
#include <list>
#include <algorithm>

using namespace std;

template<typename T>
void reverseLinkedList(list<T> &l, size_t n, size_t m)
{
    list<T>::iterator itN = l.begin();
    list<T>::iterator itM = l.begin();

    for (size_t i = 1; i < n; i++)
        itN++;

    for (size_t i = 1; i < m; i++)
        itM++;

    while (n < m)
    {
        swap(*itN++, *itM--);
        n++; m--;
    }
}

int main()
{
    list<int> l = { 1,2,3,4,5, 6 };

    reverseLinkedList(l, 2, 5);

    for (auto &e : l)
        cout << e << " ";

    return 0;
}
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　　Palindrome List
　　Given a singly linked list of characters, write a function that returns true if the given list is palindrome, else false.

#include <iostream>
#include <list>
#include <string>

using namespace std;

template<typename T>
bool palindromeList(const list<T>& l)
{
    list<T>::const_iterator it1 = l.cbegin();
    list<T>::const_reverse_iterator it2 = l.crbegin();

    for (list<T>::size_type i = 1; i < l.size() / 2; i++)
    {
        it1++; it2++;
    }

    while(it1 != l.cend()  &&  it2 != l.crend())
    {
        if (*it1++ != *it2++)
            return false;
    }

    return true;
}

int main()
{
    string str = "aabbac";
    list<char> l;

    for (auto &e : str)
        l.push_back(e);

    if (palindromeList(l))
        cout << "Yes" << endl;

    return 0;
}
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