sort-list——链表、快慢指针找中间、归并排序

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

链表，快慢指针找中点，归并排序。

注意判断条件fast->next!=NULL&&fast->next->next!=NULL，若为fast!=NULL&&fast->next!=NULL则会出现内存溢出

/**
 * 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 || head->next==NULL)
            return head;
        ListNode* slow, *fast;
        slow=head;
        fast=head;
        while(fast->next!=NULL&&fast->next->next!=NULL){
            slow=slow->next;
            fast=fast->next->next;
        }
        fast=slow;
        slow=slow->next;
        fast->next=NULL;
        fast=sortList(head);
        slow=sortList(slow);
        return merge(fast,slow);
        
    }
    ListNode *merge(ListNode *left, ListNode *right){
        ListNode *res,*temp;
        if(left==NULL){
            res=right;
            return res;
        }
        if(right==NULL){
            res=left;
            return res;
        }
        if(left->val<=right->val){
            res=left;
            left=left->next;
        }else{
            res=right;
            right=right->next;
        }
        temp=res;
        while(left!=NULL&&right!=NULL){
            if(left->val<=right->val){
                temp->next=left;
                left=left->next;
            }else{
                temp->next=right;
                right=right->next;
            }
            temp=temp->next;
            
        }
        if(left!=NULL){
            temp->next=left;
        }
        if(right!=NULL){
            temp->next=right;
        }
        return res;
    }
};