0148. Sort List (M)

Sort List (M)

题目

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

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题意

将链表排序，要求不能使用额外空间，且时间复杂度为(O(NlogN))。

思路

链表不好操作快排和堆排，使用归并排序(分治法)：每次利用快慢指针找到链表的中间位置，将其断开为左右两个子链表，待这两个子链表排序后，利用归并将它们再重新合并为一个有序链表。递归处理。

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代码实现

Java

class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        ListNode slow = head, fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode mid = slow.next;
        slow.next = null;
        return merge(sortList(head), sortList(mid));
    }

    private ListNode merge(ListNode head1, ListNode head2) {
        ListNode head = new ListNode(0);
        ListNode cur = head;
        while (head1 != null && head2 != null) {
            if (head1.val < head2.val) {
                cur.next = head1;
                head1 = head1.next;
            } else {
                cur.next = head2;
                head2 = head2.next;
            }
            cur = cur.next;
        }
        cur.next = head1 == null ? head2 : head1;
        return head.next;
    }
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var sortList = function (head) {
  if (!head || !head.next) {
    return head
  }

  let slow = head, fast = head
  while (fast.next && fast.next.next) {
    fast = fast.next.next
    slow = slow.next
  }

  let left = head, right = slow.next
  slow.next = null
  left = sortList(left)
  right = sortList(right)

  let dummy = new ListNode()
  let p = dummy
  while (left && right) {
    let node = left.val <= right.val ? left : right
    let tmp = node.next
    node.next = null
    p.next = node
    p = p.next
    node === left ? (left = tmp) : (right = tmp)
  }
  p.next = !left ? right : left
  
  return dummy.next
}