148. Sort List

问题描述：

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

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

解题思路：

说到O(nlogn)那就必然想到快速排序和归并排序和堆排序

参考大佬整理的方法

代码：

归并排序

这里用快慢指针找到中点并且然后将链表打碎成2个链表

然后对两个链表分别排序后，再调用归并。

/**
 * 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 || !head->next)
            return head;
        ListNode *slow = head, *fast = head, *pre = head;
        while(fast && fast->next){
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        pre->next = NULL;
        return merge(sortList(head), sortList(slow));
    }
    ListNode* merge(ListNode* l1, ListNode* l2){
        if(!l1) return l2;
        if(!l2) return l1;
        if(l1->val < l2->val) {
            l1->next = merge(l1->next, l2);
            return l1;
        } else {
            l2->next = merge(l1, l2->next);
            return l2;
        }
    }
};

堆排序：

注意比较器的重写！！

struct cmp{
        bool operator() (const ListNode* l1, const ListNode* l2){
            return l1->val > l2->val;
        }

参考链接：priority_queue & 结构体||类 & 自定义比较函数cmp

以及使用：

priority_queue<ListNode*, vector<ListNode*>, cmp> q;

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
    struct cmp{
        bool operator() (const ListNode* l1, const ListNode* l2){
            return l1->val > l2->val;
        }
    };
public:
    ListNode* sortList(ListNode* head) {
        if(!head || !head->next)
            return head;
        priority_queue<ListNode*, vector<ListNode*>, cmp> q;
        ListNode* ret;
        ListNode *p = head;
        while(p){
            ListNode* temp = p->next;
            p->next = NULL;
            q.push(p);
            p = temp;
        }
        ret = q.top();
        q.pop();
        ListNode *pre = ret;
        while(!q.empty()){
            ListNode *cur = q.top();
            q.pop();
            pre->next = cur;
            pre = cur;
        }
        return ret;
    }
};