🌗 148. 排序链表

难度: 🌗

问题描述

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解法

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
        // 思路：
        // 归并排序 - 快慢指针找中点
        return mySol(head);
    }

    private ListNode mySol(ListNode head) {
        // 递归终止条件
        if(head == null || head.next == null) {
            return head;
        }
        // 至少 2 个节点
        ListNode prev = getMid(head);
        ListNode left = head;
        ListNode right = prev.next;
        prev.next = null;
        ListNode a = mySol(left);
        ListNode b = mySol(right);
        return merge(a, b);
    }

    private ListNode merge(ListNode left, ListNode right) {
        ListNode res = new ListNode(-1);
        ListNode cur = res;
        while(left != null && right != null) {
            if(left.val <= right.val) {
                cur.next = left;
                left = left.next;
                cur = cur.next;
            } else {
                cur.next = right;
                right = right.next;
                cur = cur.next;
            }
        }
        while(left != null) {
            cur.next = left;
            left = left.next;
            cur = cur.next;
        }
        while(right != null) {
            cur.next = right;
            right = right.next;
            cur = cur.next;
        }
        return res.next;
    }

    private ListNode getMid(ListNode head) {
        // 找到链表偏左的那个中点的前一个节点，方便后续拆分
        ListNode slow = head;
        ListNode prev = head;
        ListNode fast = head.next;
        while(fast != null && fast.next != null) {
            fast = fast.next.next;
            prev = slow;
            slow = slow.next;
        }
        return prev;
    }
}

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输出