【LeetCode】Sort List

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Sort List

Total Accepted: 22650 Total Submissions: 110501My Submissions

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

因为要求是O(n log n)，所以选择用归并排序的方法。但因为是linked list，所以需要先找出中点的位置。可以采用两个指针，一个每次移动两步，一个每次移动一步，当快指针走到底的时候，满指针刚好走到中间。

/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode sortList(ListNode head) {        if(head==null||head.next==null) return head;        //if(head.next == null) return head;        ListNode fast = head;        ListNode slow = head;        while(fast.next!=null&&fast.next.next!=null){            fast = fast.next.next;            slow = slow.next;        }        ListNode mid = slow.next;        slow.next = null;       ListNode result =  merge(sortList(head),sortList(mid));       return result;    }    ListNode merge(ListNode left,ListNode right)    {                ListNode result = new ListNode(0);        ListNode tmp = result;        while(left!=null&&right!=null)        {            if(left.val < right.val)            {                tmp.next = left;                tmp = tmp.next;                left = left.next;            }            else            {                tmp.next = right;                tmp = tmp.next;                right = right.next;            }        }        if(left== null)        {            tmp.next = right;        }        else        {            tmp.next = left;        }                return result.next;    }}

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