题解 | #单链表的排序#

利用归并排序来求解

首先用快慢指针来找到链表的中点，把链表分为前后两部分，然后对自链表进行排序

最后把两个已排好序的链表就行合并

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 the head node
     * @return ListNode类
     */
    ListNode* merge(ListNode* head1,ListNode* head2){//合并两条排序的链表
        if(head1==NULL) return head2;
        if(head2==NULL) return head1;
        ListNode *newhead=new ListNode(0),*p=newhead;
        while(head1&&head2){
            if(head1->val<head2->val){
                p->next=head1;
                p=p->next;
                head1=head1->next;
            }else{
                p->next=head2;
                p=p->next;
                head2=head2->next;
            }
        }
        while(head1){
            p->next=head1;
            p=p->next;
            head1=head1->next;
        }
        while(head2){
            p->next=head2;
            p=p->next;
            head2=head2->next;
        }
        return newhead->next;
    }
    ListNode* sort(ListNode *head){  //归并排序
        if(head==NULL||head->next==NULL) return head;
        ListNode *fast=head,*slow=head,*pre;
        while(fast &&fast->next){  //找到中点
            fast=fast->next->next;
            pre=slow;
            slow=slow->next;
        }
        pre->next=NULL;//head--pre是前半部分
        ListNode *head1=sort(slow);//slow--NULL是后半部分
        ListNode* head2=sort(head);
        return merge(head1,head2);
    }
    ListNode* sortInList(ListNode* head) {
        // write code here
        return sort(head);
    }
};