【面经笔记】链表归并排序

链表适合归并排序
时间复杂度：O（nlogn）
空间复杂度：O（1）
稳定性：稳定

#include <iostream>using namespace std;class Node{public:    Node(){ next = NULL; val = 0; };    Node(int i){ next = NULL; val = i; };    Node* next;    int val;};Node* merge(Node* pleft, Node* pright){    if (pleft == NULL)    {        return pright;    }    if (pright == NULL)    {        return pleft;    }    Node dummy;    Node* ptmp = &dummy;    while (pleft != NULL&&pright != NULL)    {        if (pleft->val < pright->val)        {            ptmp->next = pleft;            pleft = pleft->next;            ptmp = ptmp->next;        }        else        {            ptmp->next = pright;             pright = pright->next;            ptmp = ptmp->next;        }    }    if (pleft!=NULL)    {        ptmp->next = pleft;    }    if (pright!=NULL)    {        ptmp->next = pright;    }    return dummy.next;}Node * MergeSort(Node * phead){    if (phead == NULL|| phead->next == NULL)    {        return phead;    }    Node * pfast(phead);    Node * pslow(phead);    while (pfast->next != NULL&&pfast->next->next != NULL)    //while (pfast != NULL&&pfast->next != NULL)错误，无法分割两个点    {        pfast = pfast->next->next;        pslow = pslow->next;    }    Node * pleftpart = phead;    Node * prightpart = pslow->next;    pslow->next = NULL;    pleftpart = MergeSort(pleftpart);    prightpart = MergeSort(prightpart);    return merge(pleftpart, prightpart);}int main(){    Node n1(4);    Node n2(3);    Node n3(2);    Node n4(1);    n1.next = &n2;    n2.next = &n3;    n3.next = &n4;    auto tmp = MergeSort(&n1);    while (tmp)    {        cout << tmp->val << endl;        tmp = tmp->next;    }    return 0;}

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查找中间节点使用快慢指针

while (pfast != NULL&&pfast->next != NULL) //无法分割两个点的情况