LeetCode-Sort List

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

这题的时间复杂度要求是O(n logn),很容易想到用mergeSort来解。

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
 
public class Solution {
    
    public ListNode merge(ListNode head1, ListNode head2){
        ListNode dummy = new ListNode(-1);
        ListNode tail = dummy;
        while(head1 != null && head2 != null){
            if(head1.val < head2.val){
                tail.next = head1;
                head1 = head1.next;
            }
            else {
                tail.next = head2;
                head2 = head2.next;
            }
            tail = tail.next;
        }
       　while (head1 != null){
            tail.next = head1;
　　　　　　　head1 = head1.next;
　　　　　　　tail = tail.next;

        }
        while(head2 != null){
            tail.next = head2;
　　　　　　　head2 = head2.next;
　　　　　　　tail = tail.next;

        }

        return dummy.next;
        
    }
    
    
    public ListNode findMid(ListNode head){
        ListNode slow = head;
        ListNode fast = head.next;//instead of = head to ensure slow is in the mid position
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    
    public ListNode sortList(ListNode head) {
        if(head ==null || head.next == null){
            return head;
        }
        ListNode mid = findMid(head);
        ListNode right = sortList(mid.next);
        mid.next=null;
        ListNode left = sortList(head);
        
        return  merge(left,right);
        
        
    }

    
}

其中值得注意的是，

1.在merge function里dummy前置节点和tail的作用，

2.以及findMin里ListNode fast = head.next的作用，

3.sortList中mid.next = null 的作用

 

 

二刷：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
        if(head == null || head.next == null){
            return head;
        }
        ListNode mid = findMid(head);
        ListNode right = sortList(mid.next);
        mid.next = null;
        
        ListNode left = sortList(head);
        
        return merge(right, left);
        
        
        
    }
    
    public ListNode merge (ListNode head1, ListNode head2){
        ListNode dummy =  new ListNode(-1);
        ListNode temp = dummy;
        while(head1 != null && head2 != null){
            if(head1.val < head2.val){
                temp.next = head1;
                head1 = head1.next;
            }
            else{
                temp.next = head2;
                head2 = head2.next;
                
            }
            temp = temp.next;
        }
        if(head1 != null){
            temp.next = head1;
            //head1 = head1.next;
            //temp = temp.next;
        }
        if(head2 != null){
            temp.next = head2;
            //head2 = head2.next;
            //temp = temp.next;
        }
        return dummy.next;
    }
    
    public ListNode findMid(ListNode head){
        ListNode slow = head;
        ListNode fast = head.next;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast =fast.next.next;
        }
        return slow;
        
    }
}