链表续
2019-12-26 本文已影响0人
couriravant
148. 排序链表
class Solution {
public ListNode sortList(ListNode head) {
if (head == null || head.next == null)
return head;
ListNode fast = head.next, slow = head;
//快慢指针找到中间节点
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
//从中间节点把两边断开
ListNode tmp = slow.next;
slow.next = null;
//head、temp分别是左右两个链表的头节点
ListNode left = sortList(head);
ListNode right = sortList(tmp);
//放一个临时节点把左右两个已排序链表合并
ListNode h = new ListNode(0);
ListNode res = h;
System.out.println(res.val);
while (left != null && right != null) {
if (left.val < right.val) {
h.next = left;
left = left.next;
} else {
h.next = right;
right = right.next;
}
h = h.next;
}
// left和right哪个是最后没被h.next加入的,给加入
h.next = left != null ? left : right;
System.out.println(res.val);
return res.next;
}