148. Sort List-优快云博客

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，通过快速排序算法实现，展示了具体实例及代码实现细节。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Sort a linked list in O(n log n) time using constant space complexity.

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

难度：medium

题目：排列链表，时间复杂度为O(n logn) 空间复杂度为O(1).

思路：快速排序

Runtime: 232 ms, faster than 9.03% of Java online submissions for Sort List.
Memory Usage: 41.8 MB, less than 100.00% of Java online submissions for Sort List.

/**
 * 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) {
        quickSort(head, null);
        return head;
    }
    
    private void quickSort(ListNode head, ListNode tail) {
        if (head == tail) {
            return;
        }
        int val = head.val;
        ListNode prevSwapPtr = head;
        for (ListNode ptr = head.next; ptr != tail; ptr = ptr.next) {
            if (ptr.val < val) {
                int t = prevSwapPtr.next.val;
                prevSwapPtr.next.val = ptr.val;
                ptr.val = t;
                
                prevSwapPtr = prevSwapPtr.next;
            }
        }
        head.val = prevSwapPtr.val;
        prevSwapPtr.val = val;

        quickSort(head, prevSwapPtr);
        quickSort(prevSwapPtr.next, tail);
    }
}