C语言实现单链表面试题（进阶篇）

首先给出单链表的结构，下面实现具体代码

typedef int DataType;

typedef struct Node
{
    DataType data;
    struct Node*next;
}Node,*pNode,*pList;//结点

typedef struct ComplexNode
{
    DataType D;
    struct ComplexNode*next;
    struct ComplexNode*random;
}ComplexNode,*pComplexNode;

判断单链表是否带环，若带环，求环的长度，求环的入口点

1. 判断是否带环

pNode IfRing(pList plist)//判断单链表是否带环,返回交点
{
    pNode slow = plist;
    pNode fast = plist;

    //是否带环
    while (fast&&fast->next)//注意边界问题
    {
        slow = slow->next;
        fast = fast->next->next;
        if (fast == slow)//如果相遇，则带环
        {
            return fast;
        }
    }
    return NULL;
}

1. 求换的长度

int GetCircleLen(pNode meet)//求环的长度
{

    pNode cur = meet;
    int count = 0;
    do 
    {
        count++;
        cur = cur->next;
    } while (cur!=meet);
    return count;
}

1. 求环的入口点

pNode GetCircleEntry(pList plist,pNode meet)//求环的入口点
{
    pNode cur = plist;
    pNode fast = plist;
    pNode slow = plist;
    while(cur!=meet)//根据画图可分析出
    {
        cur  = cur->next;
        meet = meet->next;
    }
    return cur;
}

判断两个链表是否相交，若相交，求交点。（假设链表不带环）

pNode CheckCrossNode(pList l1,pList l2)
{
    pNode cur1 = l1;
    pNode tail = l1;
    pNode ret = NULL;
    while(tail->next)
    {
        tail = tail->next;
    }
    tail->next = l2;
    ret = IfRing(l1);
    return ret;
}

判断两个链表是否相交，若相交，求交点。（假设链表可能带环也可能不带环）

思想：
1. 首先应判断链表是否带环：都不带环，只有一个带环
2. 都带环：入口点在环外，入口点在环内。
所有情况如下图所示

复杂链表的复制

复杂链表的构建如上面已经给出

ComplexNode CrecteComplexNode(DataType d)//这里创建复杂链表的结点

{
    pComplexNode newNode = (pComplexNode)malloc(sizeof(ComplexNode));
    if (newNode == NULL)
    {
        perror("malloc");
        return NULL;
    }
    newNode->D = d;
    newNode->next = NULL;
    newNode->random = NULL;
    return newNode;

}
void PrintComplexList(pComplexNode head)
{
    pComplexNode cur = head;
    while(cur)
    {
        printf ("%d-->",cur->D);
        printf ("random-->[%d]--->next",cur->random->D);
        cur = cur->next;
    }
    printf ("\n");
}

pComplexNode CloneComplexNode(pComplexNode head)//复制复杂链表
{
    pComplexNode cur = head;
    pComplexNode tmp = NULL;
    pComplexNode copy = NULL;
    pComplexNode tail = NULL;
    while(cur)
    {
        pComplexNode newnode = CrecteComplexNode(cur->D);
        tmp = cur;
        cur= cur->next;
        newnode->next = cur;
        tmp->next = newnode;
    }//复制每个节点并插入到节点后
    cur = head;
    while(cur)
    {
        cur->next->random = cur->random->next;
        cur = cur->next->next;
    }//调整random指针
    cur = head;
    copy= cur->next;
    tail = copy;
    while(tail->next)
    {
        tail->next = tail->next->next;
        cur->next = tail->next;
        cur = cur->next;
        tail = tail->next;
        cur->next = NULL;
        return copy;
    }
}

//下面给出具体的测试代码

void test3()
{
    pComplexNode pNode1 = CrecteComplexNode(1);
    pComplexNode pNode2 = CrecteComplexNode(2);
    pComplexNode pNode3 = CrecteComplexNode(3);
    pComplexNode pNode4 = CrecteComplexNode(4);
    pComplexNode pNode5 = CrecteComplexNode(5);


    pNode1->next = pNode2;
    pNode2->next = pNode3;
    pNode3->next = pNode4;

    pNode1->random = pNode4;
    pNode2->random = pNode1;
    pNode3->random = pNode2;
    pNode4->random = pNode2;
    CloneComplexNode(pNode1);
    PrintComplexList(pNode1);

}

这些是c语言实现链表的面试题中较为复杂的了，链表重要的就是逻辑，把主要过程理解清楚。