148. Sort List

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

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null)return head;

        final ListNode middle = findMiddle(head);
        final ListNode head2 = middle.next;
        middle.next = null; // 断开

        final ListNode l1 = sortList(head);  // 前半段排序
        final ListNode l2 = sortList(head2);  // 后半段排序
        return mergeTwoLists(l1, l2);
    }

    // Merge Two Sorted Lists
    private static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        for (ListNode p = dummy; l1 != null || l2 != null; p = p.next) {
            int val1 = l1 == null ? Integer.MAX_VALUE : l1.val;
            int val2 = l2 == null ? Integer.MAX_VALUE : l2.val;
            if (val1 <= val2) {
                p.next = l1;
                l1 = l1.next;
            } else {
                p.next = l2;
                l2 = l2.next;
            }
        }
        return dummy.next;
    }
    // 快慢指针找到中间节点
    private static ListNode findMiddle(ListNode head) {
        if (head == null) return null;

        ListNode slow = head;
        ListNode fast = head.next;

        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
}

这种答案就是让人看着非常的舒适，有一种为什么他妈的这么好的感觉，嗯

上面是2019春写的，当时批话真的多还菜，当然现在也没改变多少--2019冬

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(left, right);
    }
    public ListNode findmid(ListNode head){
        ListNode slow = head, fast = head.next;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    public ListNode merge(ListNode l1, ListNode l2){
        ListNode dummy = new ListNode(-1);
        ListNode tail = dummy;
        while(l1 != null && l2 != null){
            if(l1.val < l2.val){
                tail.next = l1;
                l1 = l1.next;
            }
            else{
                tail.next = l2;
                l2 = l2.next;
            }
            tail = tail.next;
        }
        tail.next = l1 != null ? l1 : l2;
        return dummy.next;
    }
}