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Sort List (使用归并排序的链表排序)

更新时间:2022-08-30 18:13:40

Sort a linked list in O(n log n) time using constant space complexity.

 

C++代码的实现:

#include<iostream>
#include<new>
using namespace std;

//Definition for singly-linked list.
struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};
class Solution
{
public:
    ListNode *sortList(ListNode *head)
    {
        if(head==NULL)
            return NULL;
        //递归的结束条件,注意,与对数组的归并排序不同
        if(head->next==NULL)
            return head;
        ListNode *p=head;
    //q不能从head开始,为什么呢? ListNode
*q=head->next; ListNode *l2=NULL; while(q&&q->next) { p=p->next; q=q->next->next; } l2=p->next; p->next=NULL; return Merge(sortList(head),sortList(l2)); } ListNode *Merge(ListNode *l1,ListNode *l2) { ListNode *pre=l1; ListNode *p=l1; ListNode *q=l2; while(p&&q) { if(p->val<=q->val) { pre=p; p=p->next; continue; } else { l2=q->next; q->next=NULL; q->next=p; if(p==l1) l1=q; else pre->next=q; pre=q; q=l2; } } if(q) pre->next=q; return l1; } void createList(ListNode *&head) { ListNode *p=NULL; int i=0; int arr[10]= {5,7,6,4,8,3,9,3,10,1}; for(i=0; i<10; i++) { if(head==NULL) { head=new ListNode(arr[i]); if(head==NULL) return; } else { p=new ListNode(arr[i]); p->next=head; head=p; } } } }; int main() { Solution s; ListNode *L=NULL; s.createList(L); ListNode *L1=NULL; L1=s.sortList(L); while(L1) { cout<<L1->val<<" "; L1=L1->next; } }

运行结果:

Sort List (使用归并排序的链表排序)

参考:http://www.tuicool.com/articles/2eemi2

http://www.cnblogs.com/tenosdoit/p/3666585.html