148. Sort List
Sort a linked list in O(n log n) time using constant space complexity.
解法
归并排序。用快慢指针法找到中间结点。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode mid = findMiddle(head);
ListNode right = sortList(mid.next);
mid.next = null;
ListNode left = sortList(head);
return merge(left, right);
}
private ListNode findMiddle (ListNode head) {
ListNode slow = head;
ListNode fast = head.next;
while (fast!= null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
private ListNode merge(ListNode left, ListNode right) {
ListNode dummy = new ListNode(0);
ListNode head = dummy;
while (left != null && right != null) {
if (left.val < right.val) {
head.next = left;
left = left.next;
} else {
head.next = right;
right = right.next;
}
head = head.next;
}
if (left != null) {
head.next = left;
}
if (right != null) {
head.next = right;
}
return dummy.next;
}
}
本文介绍了一种在O(nlogn)时间内使用常数空间复杂度对链表进行排序的方法。采用归并排序策略,利用快慢指针寻找中点,并递归地将链表分为两部分进行排序。
2353

被折叠的 条评论
为什么被折叠?



