0.目錄

1.連結串列的基本操作

• 1.1 結點定義
• 1.2 建立連結串列
• 1.3 銷燬連結串列
• 1.4 列印連結串列
• 1.5 獲取連結串列長度

2.結點的基本操作

• 2.1 刪除結點
• 2.2 查詢結點

3.面試題

• 3.1 反轉連結串列
• 3.2 合併兩個單向排序連結串列
• 3.3 查詢兩個連結串列的第一個公共結點

1.連結串列的基本操作

1.1 結點定義

#include <iostream>

using namespace std;

struct Node
{
    int value;
    Node* next;
};
 

1.2 建立連結串列

Node* createLinkedList(int data[], int len)
{
    Node* ret = NULL;
    Node* slider = NULL;

    for(int i=0; i<len; i++)
    {
        Node* n = new Node();

        n->value = data[i];
        n->next = NULL;

        if( slider == NULL )
        {
            slider = n;
            ret = n;
        }
        else
        {
            slider->next = n;
            slider = n;
        }
    }

    return ret;
}
 

1.3 銷燬連結串列

void destroyLinkedList(Node* list)
{
    while( list )
    {
        Node* del = list;

        list = list->next;

        delete del;
    }
}

1.4 列印連結串列

void printLinkedList(Node* list)
{
    while( list )
    {
        cout << list->value << "->";

        list = list->next;
    }

    cout << "NULL" << endl;
}
 

1.5 獲取連結串列長度

int getListLength(Node* list)
{
    int ret = 0;

    while( list )
    {
        ret++;
        list = list->next;
    }

    return ret;
}

測試：

int main()
{
    int a[] = {1, 5, 3, 2, 4};

    Node* list1 = createLinkedList(a, 5);
    printLinkedList(list1);
    cout << getListLength(list1) << endl;
    destroyLinkedList(list1);
    cout << endl;

    Node* list2 = createLinkedList(NULL, 0);
    printLinkedList(list2);
    cout << getListLength(list2) << endl;
    destroyLinkedList(list2);
    cout << endl;

    int b[] = {6};
    Node* list3 = createLinkedList(b, 1);
    printLinkedList(list3);
    cout << getListLength(list3) << endl;
    destroyLinkedList(list3);

    return 0;
}

執行結果為：

1->5->3->2->4->NULL
5

NULL
0

6->NULL
1

2.結點的基本操作

2.1 刪除結點

Node* deleteNode(Node* list, int value)
{
    Node* head = list;
    Node* slider = NULL;

    while( head && (head->value == value) )
    {
        slider = head;
        head = head->next;
        slider = NULL;
    }

    Node* ret = head;

    while( ret )
    {
        slider = ret->next;

        if( slider && (slider->value == value) )
        {
            ret->next = slider->next;
            slider = NULL;
        }
        else
        {
            ret = ret->next;
        }
    }

    return head;
}

測試：

int main()
{
    int a[] = {1, 2, 3, 2, 5};

    Node* list1 = createLinkedList(a, 5);
    printLinkedList(list1);
    printLinkedList(deleteNode(list1, 2));
    destroyLinkedList(list1);
    cout << endl;

    Node* list2 = createLinkedList(NULL, 0);
    printLinkedList(list2);
    printLinkedList(deleteNode(list2, 2));
    destroyLinkedList(list2);
    cout << endl;

    int b[] = {2, 2, 2, 2, 2};
    Node* list3 = createLinkedList(b, 5);
    printLinkedList(list3);
    printLinkedList(deleteNode(list3, 2));
    destroyLinkedList(list3);
    cout << endl;

    int c[] = {1};
    Node* list4 = createLinkedList(c, 1);
    printLinkedList(list4);
    printLinkedList(deleteNode(list4, 2));
    destroyLinkedList(list4);

    return 0;
}

執行結果為：

1->2->3->2->5->NULL
1->3->5->NULL

NULL
NULL

2->2->2->2->2->NULL
NULL

1->NULL
1->NULL

2.2 查詢結點

Node* findNode(Node* list, int value)
{
    Node* ret = NULL;
    Node* slider = list;

    while( slider )
    {
        if( slider->value == value )
        {
            ret = slider;
            break;
        }
        else
        {
            slider = slider->next;
        }
    }

    return ret;
}

3.面試題

3.1 反轉連結串列

Node* reverseLinkedList(Node* list)
{
    Node* ret = NULL;
    Node* slider = list;
    Node* next = NULL;

    while( slider )
    {
        next = slider->next;
        slider->next = ret;
        ret = slider;
        slider = next;
    }

    return ret;
}

測試：

int main()
{
    int a[] = {1, 5, 3, 2, 4};

    Node* list1 = createLinkedList(a, 5);
    printLinkedList(list1);
    printLinkedList(reverseLinkedList(list1));
    destroyLinkedList(list1);
    cout << endl;

    Node* list2 = createLinkedList(NULL, 0);
    printLinkedList(list2);
    printLinkedList(reverseLinkedList(list2));
    destroyLinkedList(list2);
    cout << endl;

    int b[] = {6};
    Node* list3 = createLinkedList(b, 1);
    printLinkedList(list3);
    printLinkedList(reverseLinkedList(list3));
    destroyLinkedList(list3);

    return 0;
}

執行結果為：

1->5->3->2->4->NULL
4->2->3->5->1->NULL

NULL
NULL

6->NULL
6->NULL

3.2 合併兩個單向排序連結串列

Node* mergeLinkedList(Node* list1, Node* list2)
{
    Node* ret = NULL;

    if( list1 == NULL )
    {
        ret = list2;
    }
    else if( list2 == NULL )
    {
        ret = list1;
    }
    else
    {
        if( list1->value < list2->value )
        {
            ret = list1;
            list1 = list1->next;
        }
        else
        {
            ret = list2;
            list2 = list2->next;
        }

        Node* slider = ret;

        while( list1 && list2 )
        {
            if( list1->value < list2->value )
            {
                slider->next = list1;
                list1 = list1->next;
            }
            else
            {
                slider->next = list2;
                list2 = list2->next;
            }

            slider = slider->next;
        }

        if( list1 == NULL )
        {
            slider->next = list2;
        }
        else if( list2 == NULL )
        {
            slider->next = list1;
        }
    }

    return ret;
}

測試：

int main()
{
    int a[] = {1, 2, 4, 6, 8};

    Node* list1 = createLinkedList(a, 5);
    printLinkedList(list1);

    int b[] = {2, 2, 3, 3, 7};
    Node* list2 = createLinkedList(b, 5);
    printLinkedList(list2);

    Node* list3 = mergeLinkedList(list1, list2);
    printLinkedList(list3);
    destroyLinkedList(list3);

    return 0;
}

執行結果為：

1->2->4->6->8->NULL
2->2->3->3->7->NULL
1->2->2->2->3->3->4->6->7->8->NULL

3.3 查詢兩個連結串列的第一個公共結點

Node* findFirstCommonNode(Node* list1, Node* list2)
{
    int len1 = getListLength(list1);
    int len2 = getListLength(list2);
    Node* ret = NULL;

    if( len1 > len2 )
    {
        for(int i=0; i<(len1-len2); i++)
        {
            list1 = list1->next;
        }
    }
    else
    {
        for(int i=0; i<(len2-len1); i++)
        {
            list2 = list2->next;
        }
    }

    while( list1 )
    {
        if( list1 == list2 )
        {
            ret = list1;
            break;
        }
        else
        {
            list1 = list1->next;
            list2 = list2->next;
        }
    }

    return ret;
}

測試：

int main()
{
    int a[] = {1, 2, 3};
    Node* list1 = createLinkedList(a, 3);

    int b[] = {4, 5};
    Node* list2 = createLinkedList(b, 2);

    int c[] = {6, 7};
    Node* list3 = createLinkedList(c, 2);

    Node* ret = NULL;

    ret = list1;
    while( ret->next )
    {
        ret = ret->next;
    }
    ret->next = list3;

    ret = list2;
    while( ret->next )
    {
        ret = ret->next;
    }
    ret->next = list3;

    printLinkedList(list1);
    printLinkedList(list2);

    ret = findFirstCommonNode(list1, list2);
    printLinkedList(ret);

    return 0;
}

執行結果為：

1->2->3->6->7->NULL
4->5->6->7->NULL
6->7->NULL