[LeetCode] 148. Sort List
Given the head of a linked list, return the list after sorting it in ascending order.
Follow up: Can you sort the linked list in O(n logn) time and O(1) memory (i.e. constant space)?
Example 1:
Input: head = [4,2,1,3] Output: [1,2,3,4]
Example 2:
Input: head = [-1,5,3,4,0] Output: [-1,0,3,4,5]
Example 3:
Input: head = [] Output: []
Constraints:
- The number of nodes in the list is in the range
[0, 5 * 104]. -105 <= Node.val <= 105
Follow up: Can you sort the linked list in O(n logn) time and O(1) memory (i.e. constant space)?
排序链表。
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。这道题的 followup 是满足时间 O(nlogn)，空间 O(1) 的要求。
按照题目要求，因为时间是 nlogn，所以思路只能是 merge sort。思路是找到链表的中点，然后用 merge sort 的思路递归再把链表一点点拼凑回去。
时间O(nlogn)
空间O(n) - 递归使用的栈空间
JavaScript实现
1 /** 2 * @param {ListNode} head 3 * @return {ListNode} 4 */ 5 var sortList = function(head) { 6 // corner case 7 if (head === null || head.next === null) { 8 return head; 9 } 10 let middle = findMiddle(head); 11 let next = middle.next; 12 middle.next = null; 13 return merge(sortList(head), sortList(next)); 14 }; 15 16 var findMiddle = function(head) { 17 let slow = head; 18 let fast = head; 19 while (fast.next !== null && fast.next.next !== null) { 20 slow = slow.next; 21 fast = fast.next.next; 22 } 23 return slow; 24 }; 25 26 var merge = function(a, b) { 27 let dummy = new ListNode(0); 28 let cur = dummy; 29 while (a !== null && b !== null) { 30 if (a.val < b.val) { 31 cur.next = a; 32 a = a.next; 33 } else { 34 cur.next = b; 35 b = b.next; 36 } 37 cur = cur.next; 38 } 39 if (a === null) cur.next = b; 40 else cur.next = a; 41 return dummy.next; 42 };
Java实现
1 class Solution { 2 public ListNode sortList(ListNode head) { 3 // corner case 4 if (head == null || head.next == null) { 5 return head; 6 } 7 8 // normal case 9 ListNode middle = getMiddle(head); 10 ListNode next = middle.next; 11 middle.next = null; 12 return merge(sortList(head), sortList(next)); 13 } 14 15 private ListNode getMiddle(ListNode head) { 16 ListNode slow = head; 17 ListNode fast = head; 18 while (fast.next != null && fast.next.next != null) { 19 slow = slow.next; 20 fast = fast.next.next; 21 } 22 return slow; 23 } 24 25 private ListNode merge(ListNode a, ListNode b) { 26 ListNode dummy = new ListNode(0); 27 ListNode cur = dummy; 28 while (a != null && b != null) { 29 if (a.val < b.val) { 30 cur.next = a; 31 a = a.next; 32 } else { 33 cur.next = b; 34 b = b.next; 35 } 36 cur = cur.next; 37 } 38 if (a == null) { 39 cur.next = b; 40 } 41 if (b == null) { 42 cur.next = a; 43 } 44 return dummy.next; 45 } 46 }
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