排序链表

2020-11-21

给你链表的头结点 head ，请将其按 升序 排列并返回 排序后的链表 。

进阶：
你可以在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序吗？
 

示例 1：
输入：head = [4,2,1,3]
输出：[1,2,3,4]

示例 2：
输入：head = [-1,5,3,4,0]
输出：[-1,0,3,4,5]

示例 3：
输入：head = []
输出：[]
 
提示：
链表中节点的数目在范围 [0, 5 * 104] 内
-105 <= Node.val <= 105

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/sort-list
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

插入排序

时间复杂度不符合要求

/**
 * 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) {
        if (head == null) {
            return head;
        }
        ListNode dummyHead = new ListNode(0);
        dummyHead.next = head;
        ListNode lastSorted = head, curr = head.next;
        while (curr != null) {
            if (lastSorted.val <= curr.val) {
                lastSorted = lastSorted.next;
            } else {
                ListNode prev = dummyHead;
                while (prev.next.val <= curr.val) {
                    prev = prev.next;
                }
                lastSorted.next = curr.next;
                curr.next = prev.next;
                prev.next = curr;
            }
            curr = lastSorted.next;
        }
        return dummyHead.next;
    }
}

自顶向下的归并排序

空间复杂度不符合要求

class Solution {
    public ListNode sortList(ListNode head) {
        return sortList(head, null);
    }

    public ListNode sortList(ListNode head, ListNode tail) {        //递归划分与排序
        if (head == null) {
            return head;
        }
        if (head.next == tail) {
            head.next = null;
            return head;
        }
        ListNode slow = head, fast = head;                          //两个指针，慢指针一次走一步，快指针一次走两步，等到快指针走到连边的尾部，慢指针走到链表的中间
        while (fast != tail) {
            slow = slow.next;
            fast = fast.next;
            if (fast != tail) {
                fast = fast.next;
            }
        }
        ListNode mid = slow;                                        //慢指针走到中间
        ListNode list1 = sortList(head, mid);                       //对左半部分链表划分排序
        ListNode list2 = sortList(mid, tail);                       //对右半部分链表划分排序
        ListNode sorted = merge(list1, list2);                      //合并两个子链表
        return sorted;
    }

    public ListNode merge(ListNode head1, ListNode head2) {
        ListNode dummyHead = new ListNode(0);
        ListNode temp = dummyHead, temp1 = head1, temp2 = head2;
        while (temp1 != null && temp2 != null) {                    //两个子链表都没有遍历到底时
            if (temp1.val <= temp2.val) {                           //如果第一个链表的当前值小于第二个链表的当前值，有序链表加入第一个链表的当前值
                temp.next = temp1;
                temp1 = temp1.next;
            } else {                                                //有序链表加入第二个链表的当前值
                temp.next = temp2;
                temp2 = temp2.next;
            }
            temp = temp.next;
        }
        //将多出的一个加入到链表末尾
        if (temp1 != null) {
            temp.next = temp1;
        } else if (temp2 != null) {
            temp.next = temp2;
        }
        return dummyHead.next;
    }
}

自底向上的归并排序


class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null) {
            return head;
        }
        int length = 0;
        ListNode node = head;
        while (node != null) {                                              // 统计链表的长度
            length++;
            node = node.next;
        }

        ListNode dummyHead = new ListNode(0, head);
        for (int subLength = 1; subLength < length; subLength <<= 1) {      // subLength <<= 1是sublength*2
            ListNode prev = dummyHead, curr = dummyHead.next;
            while (curr != null) {
                ListNode head1 = curr;
                for (int i = 1; i < subLength && curr.next != null; i++) {  // 划分subLength长度
                    curr = curr.next;
                }
                ListNode head2 = curr.next;
                curr.next = null;
                curr = head2;
                for (int i = 1; i < subLength && curr != null && curr.next != null; i++) {// 在上个基础上划分subLength长度
                    curr = curr.next;
                }
                ListNode next = null;
                if (curr != null) {                                                       //对剩余的排序
                    next = curr.next;
                    curr.next = null;
                }
                ListNode merged = merge(head1, head2);                                    //归并排序两个链表
                prev.next = merged;
                while (prev.next != null) {
                    prev = prev.next;
                }
                curr = next;
            }
        }
        return dummyHead.next;
    }

    public ListNode merge(ListNode head1, ListNode head2) {                              //归并两个链表
        ListNode dummyHead = new ListNode(0);
        ListNode temp = dummyHead, temp1 = head1, temp2 = head2;
        while (temp1 != null && temp2 != null) {
            if (temp1.val <= temp2.val) {
                temp.next = temp1;
                temp1 = temp1.next;
            } else {
                temp.next = temp2;
                temp2 = temp2.next;
            }
            temp = temp.next;
        }
        if (temp1 != null) {
            temp.next = temp1;
        } else if (temp2 != null) {
            temp.next = temp2;
        }
        return dummyHead.next;
    }
}

参考文献

作者：LeetCode-Solution来源：力扣（LeetCode）著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/sort-list/solution/pai-xu-lian-biao-by-leetcode-solution/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。