问题描述
Sort a linked list in O(n log n) time using constant space complexity.地址
问题分析
用归并排序解答，因为归并排序更容易操纵链表但要求用固定空间，所以必须是归并排序的非递归形式吧
经验教训
如何利用快慢指针找到链表的中间位置，并将它一分为二
代码实现
递归版
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode
sortList(ListNode head) {
if (head ==
null || head.next ==
null) {
return head;
}
ListNode slow = head;
ListNode fast = head;
ListNode tail1 =
null;
while (fast !=
null && fast.next !=
null) {
tail1 = slow;
slow = slow.next;
fast = fast.next.next;
}
if (tail1 !=
null) {
tail1.next =
null;
}
ListNode l1 = sortList(head);
ListNode l2 = sortList(slow);
return mergeSortedList(l1, l2);
}
public ListNode
mergeSortedList(ListNode l1, ListNode l2) {
ListNode dummy =
new ListNode(
0);
ListNode tail = dummy;
while (l1 !=
null && l2 !=
null) {
if (l1.val < l2.val) {
tail.next = l1;
l1 = l1.next;
}
else {
tail.next = l2;
l2 = l2.next;
}
tail = tail.next;
}
tail.next = l1 !=
null ? l1 : l2;
return dummy.next;
}
}
非递归版（以后补充）
