Leetcode 148. 排序链表 归并排序

原题链接：Leetcode 148. 排序链表

方法一：自顶向下归并排序

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* merge2Lists(ListNode* head1,ListNode* head2){if(!head1) return head2;if(!head2) return head1;if(head1-> val <=  head2->val){head1->next = merge2Lists(head1->next,head2);return head1;}else {head2->next = merge2Lists(head1,head2->next);return head2;}}ListNode* sortList(ListNode* head) {if(head==nullptr || head->next==nullptr) return head;// 快慢指针找到中间节点ListNode * slow = head;ListNode * fast = head;ListNode * pre = slow;while(fast && fast->next){pre = slow;slow = slow->next;fast = fast->next->next;}// 找到中间节点 slow，断开slow的前一个节点pre和slow的关系,将链表拆分成两个子链表pre ->next = nullptr;// 对两个子链表分别排序head = sortList(head);slow = sortList(slow);// 合并两个有序链表return merge2Lists(head,slow);}
};

方法二：自底向上归并排序

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* merge2Lists(ListNode* head1,ListNode* head2){if(!head1) return head2;if(!head2) return head1;if(head1-> val <=  head2->val){head1->next = merge2Lists(head1->next,head2);return head1;}else {head2->next = merge2Lists(head1,head2->next);return head2;}}ListNode* sortList(ListNode* head) {if(head==nullptr || head->next==nullptr) return head;int n=0;ListNode* cur = head;while(cur){cur=cur->next;n++;}ListNode* root = new ListNode (0,head);for(int len=1;len<n;len*=2){ListNode* pre = root;cur = root->next;while(cur!=nullptr){ListNode* head1 = cur;for(int i=1;i<len && cur->next!=nullptr;i++){cur = cur->next;}ListNode* head2 = cur->next;cur->next = nullptr;cur= head2;for(int i=1;i<len && cur!=nullptr && cur->next!=nullptr ;i++){cur = cur->next;}ListNode* nxt = nullptr;if(cur!=nullptr){nxt = cur->next;cur->next = nullptr;}ListNode* merged = merge2Lists(head1,head2);pre->next=merged;while(pre->next!=nullptr){pre = pre->next;}cur = nxt;}}return root->next;}
};