(java)(leecode)148. 排序链表

148. 排序链表

Src: https://leetcode-cn.com/problems/sort-list/
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？

方法1：自上而下的归并算法

时间复杂度：O(nlogn)
空间复杂度：O(logn)

执行用时：6 ms, 在所有 Java 提交中击败了53.08%的用户
内存消耗：47 MB, 在所有 Java 提交中击败了6.29%的用户

class Solution {
        public ListNode sortList(ListNode head) {
            // 链表为空或者链表为只有一个节点的时候结束递归
            if (head == null || head.next == null) {
                return head;
            }
            ListNode fast = head.next.next, slow = head, mid;
            while (fast != null&&fast.next!=null) {
                fast = fast.next.next;
                slow = slow.next;
            }
            mid = slow.next;
            slow.next = null;
            ListNode list1 = sortList(head);
            ListNode list2 = sortList(mid);
            ListNode sorted = merge(list1, list2);
            return sorted;
        }

        public ListNode merge(ListNode head1, ListNode head2) {
            ListNode head = new ListNode(0);
            ListNode temp0 = head;
            ListNode temp1 = head1, temp2 = head2;
            while (temp1 != null && temp2 != null) {
                if (temp1.val < temp2.val) {
                    temp0.next = temp1;
                    temp1 = temp1.next;
                } else {
                    temp0.next = temp2;
                    temp2 = temp2.next;
                }
                temp0 = temp0.next;
            }
            // 如果有一条是空的，就把非空的接上
            // 如果两条皆空则接上空指针
            if (temp1 == null) {
                temp1 = temp2;
            }
            temp0.next = temp1;
            return head.next;
        }
    }

方法二：自底向上归并排序

执行用时：
10 ms, 在所有 Java 提交中击败了22.57%的用户
内存消耗：46.5 MB, 在所有 Java 提交中击败了22.42%的用户

class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null) {
            return head;
        }
        int length = 0;
        ListNode node = head;
        while (node != null) {
            length++;
            node = node.next;
        }
        ListNode dummyHead = new ListNode(0, head);
        for (int subLength = 1; subLength < length; subLength <<= 1) {
            ListNode prev = dummyHead, curr = dummyHead.next;
            while (curr != null) {
                ListNode head1 = curr;
                for (int i = 1; i < subLength && curr.next != null; i++) {
                    curr = curr.next;
                }
                ListNode head2 = curr.next;
                curr.next = null;
                curr = head2;
                for (int i = 1; i < subLength && curr != null && curr.next != null; i++) {
                    curr = curr.next;
                }
                ListNode next = null;
                if (curr != null) {
                    next = curr.next;
                    curr.next = null;
                }
                ListNode merged = merge(head1, head2);
                prev.next = merged;
                while (prev.next != null) {
                    prev = prev.next;
                }
                curr = next;
            }
        }
        return dummyHead.next;
    }

    public ListNode merge(ListNode head1, ListNode head2) {
        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;
    }
}

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）
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