LeetCode刷题笔录Sort List

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

挺有意思的题，这里用了merge sort，可以利用到merge sorted list 那题的代码

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if(head == null || head.next == null)
            return head;
        ListNode slow = head, fast = head;
        //use two-pointer method to find the middle element of the list
        while(fast.next != null && fast.next.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode head2 = slow.next;
        slow.next = null;
        //recursively sorted the two lists
        head = sortList(head);
        head2 = sortList(head2);
        return merge(head, head2);
    }
    
    private ListNode merge(ListNode l1, ListNode l2){
        if(l1 == null)
            return l2;
        if(l2 == null)
            return l1;
        ListNode dummy = new ListNode(0);
        ListNode last = dummy;
        while(l1 != null && l2 != null){
            if(l1.val <= l2.val){
                last.next = l1;
                l1 = l1.next;
                last = last.next;
            }
            else{
                last.next = l2;
                l2 = l2.next;
                last = last.next;
            }
        }
        if(l1 != null)
            last.next = l1;
        else if(l2 != null)
            last.next = l2;
        return dummy.next;
    }
}