Sort List
描述
Sort a linked list in O(n log n) time using constant space complexity.
分析
常数空间且O(nlogn)，单链表适合用归并排序，双向链表适合用快速排序。本题可以复用 Merge Two Sorted Lists 的代码。
代码
// Sort List
// 归并排序，时间复杂度O(nlogn)，空间复杂度O(1)
public class Solution {
public ListNode sortList(ListNode head) {
if (head == null || head.next == null)return head;
final ListNode middle = findMiddle(head);
final ListNode head2 = middle.next;
middle.next = null; // 断开
final ListNode l1 = sortList(head); // 前半段排序
final ListNode l2 = sortList(head2); // 后半段排序
return mergeTwoLists(l1, l2);
}
// Merge Two Sorted Lists
private static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(-1);
for (ListNode p = dummy; l1 != null || l2 != null; p = p.next) {
int val1 = l1 == null ? Integer.MAX_VALUE : l1.val;
int val2 = l2 == null ? Integer.MAX_VALUE : l2.val;
if (val1 <= val2) {
p.next = l1;
l1 = l1.next;
} else {
p.next = l2;
l2 = l2.next;
}
}
return dummy.next;
}
// 快慢指针找到中间节点
private static ListNode findMiddle(ListNode head) {
if (head == null) return null;
ListNode slow = head;
ListNode fast = head.next;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
}