Sort a linked list in O(n log n) time using constant space complexity.
nlogn的排序有快速排序、归并排序、堆排序。双向链表用快排比较适合，堆排序也可以用于链表，单向链表适合用归并排序。以下是用归并排序的代码：
1 /** 2 * Definition for singly-linked list. 3 * class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 13 14 /** 15 * Definition for singly-linked list. 16 * struct ListNode { 17 * int val; 18 * ListNode *next; 19 * ListNode(int x) : val(x), next(NULL) {} 20 * }; 21 */ 22 23 public class Solution { 24 public ListNode sortList(ListNode head) { 25 if(head == null || head.next == null) return head; 26 ListNode fast = head, slow = head; 27 while(fast.next != null && fast.next.next != null){ 28 fast = fast.next.next; 29 slow = slow.next; 30 } 31 fast = slow.next; 32 slow.next = null; 33 fast = sortList(fast);// 后半段 34 slow = sortList(head);//前半段 35 return merge(slow, fast); 36 } 37 ListNode merge(ListNode head1, ListNode head2){ 38 if(head1 == null)return head2; 39 if(head2 == null)return head1; 40 ListNode ret = null, cur = null ; 41 if(head1.val < head2.val){ 42 ret = head1; 43 head1 = head1.next; 44 }else{ 45 ret = head2; 46 head2 = head2.next; 47 } 48 cur = ret; 49 while(head1 != null && head2 != null){ 50 if(head1.val < head2.val){ 51 cur.next = head1; 52 head1 = head1.next; 53 }else{ 54 cur.next = head2; 55 head2 = head2.next; 56 } 57 cur = cur.next; 58 } 59 if(head1 != null) cur.next = head1; 60 if(head2 != null) cur.next = head2; 61 return ret; 62 } 63 }
第二遍：
这次做的时候 有一句话不一样。while(fast != null && fast.next != null) 在最开始分组时， 结果导致runtime error。
Last executed input: {2,1}