148. Sort List

https://leetcode.com/problems/sort-list/description/

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

链表类题目找中间点使用双指针算法，要判断head -> next != nullptr,一定要首先判断head != nullptr。

ListNode* findMid(ListNode* head){
        ListNode* fast = head -> next;
    ListNode* slow = head;
    while(fast != nullptr && fast -> next != nullptr){
        fast = fast -> next -> next;
        slow = slow -> next;
    }
       return slow;
}

思考归并排序算法，

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* merge(ListNode* left,ListNode* right){
        
        ListNode* dummy = new ListNode(-1);
        ListNode* p = dummy;
        while(left != nullptr && right != nullptr){
            if(left -> val < right -> val){
                dummy -> next = left;
                left = left -> next;
                dummy = dummy -> next;
            }
            else{
                dummy -> next = right;
                right = right -> next;
                dummy = dummy -> next;
            }
            
        }
        if(left != nullptr){
            dummy -> next = left;
        }
        if(right != nullptr){
            dummy -> next = right;
        }
        return p -> next;
    }
    ListNode* findMid(ListNode* head){
        ListNode* fast = head -> next;
        ListNode* slow = head;
        while(fast != nullptr && fast -> next != nullptr){
            fast = fast -> next -> next;
            slow = slow -> next;
        }
        return slow;
    }
    ListNode* mergeSort(ListNode* head){
        if(head == nullptr || head -> next == nullptr){
            return head;
        }
        
        ListNode* midNode = findMid(head);        
        ListNode* left = mergeSort(midNode -> next);
        midNode -> next = nullptr;
        ListNode* right = mergeSort(head);
        left = merge(left,right);
        return left;
        
        
    }
    ListNode* sortList(ListNode* head) {
        if(head == nullptr){
            return head;
        }        
        return mergeSort(head);
    }
};