头条重点
题目
在 O(n log n) 时间复杂度和常数级空间复杂度下,对链表进行排序。
示例 1:
输入: 4->2->1->3
输出: 1->2->3->4
示例 2:
输入: -1->5->3->4->0
输出: -1->0->3->4->5
解题思路
- 通过快慢指针将链表拆分
- 递归进行拆分,再通过合并两个排序链表的方式进行合并
- 类似于归并排序
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) {
fast = fast.next.next;
slow = slow.next;
}
ListNode mid = slow.next;
slow.next = null;
ListNode l1 = sortList(head);
ListNode l2 = sortList(mid);
return merge(l1, l2);
}
private ListNode merge(ListNode l1, ListNode l2) {
if (l1 == null) {
return l2;
}
if (l2 == null) {
return l1;
}
ListNode head,res;
if (l1.val > l2.val) {
head = l2;
l2 = l2.next;
} else {
head = l1;
l1 = l1.next;
}
res = head;
// head.next = null;
while (l1 != null || l2 != null) {
if (l1 == null) {
head.next = l2;
l2 = l2.next;
} else if (l2 == null) {
head.next = l1;
l1 = l1.next;
} else {
if (l1.val > l2.val) {
head.next = l2;
l2 = l2.next;
} else {
head.next = l1;
l1 = l1.next;
}
}
head = head.next;
}
return res;
}