148. Sort List 链表归并排序

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/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */void mergeLink(ListNode* head,ListNode* mid, ListNode* last){    ListNode* head1=head;    ListNode* end1 = mid;    ListNode* rehead2 = mid->next;//记录第一个链表的结尾    ListNode* head2 = mid->next,*end2 = last;    ListNode* temp1,*temp2;    while(head1->next!=rehead2){        while(head1->next!=rehead2 && head1->next->val<=head2->val){//当第一个链表小雨第二个链表就一直遍历            head1=head1->next;        }        if(head1->next == rehead2){//当第一个链表到尾了直接把第二个链表接到第一个链表上            head1->next = head2;            break;        }else{            temp2 = head2->next;            temp1 = head1->next;            head1->next = head2;            head2->next = temp1;            head2 = temp2;            head1=head1->next;            if(head2 == end2){//当第二个链表到尾了，找到第一个链表的尾然后把第二个链表的尾接到第一个链表的尾                while(head1->next!=rehead2) head1 = head1->next;                head1->next = head2;                break;            }        }    }} void mergesort(ListNode* head,ListNode* last){    ListNode* fast = head,*slow = head;    if(head->next == last || head->next->next==last) return;    while(fast!=last && fast->next!=last){//寻找中位数        fast=fast->next->next;        slow = slow->next;    }    mergesort(head,slow->next);    mergesort(slow,last);    mergeLink(head,slow,last);//合并两个链表    return;}class Solution {public:    ListNode* sortList(ListNode* head) {        if(head==NULL || head->next == NULL) return head;        ListNode* newHead = new ListNode(0);        newHead->next = head;        mergesort(newHead,nullptr);        return newHead->next;    }};
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