148. Sort List
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Sort a linked list in O(n log n) time using constant space complexity.
这是用迭代的方法进行归并排序 把链表拆成两个 然后分别排序 再融合成一个
/** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { //use the fast and slow pointer to get the middle of a ListNode ListNode getMiddleOfList(ListNode head) { ListNode slow = head; ListNode fast = head; while(fast.next!=null&&fast.next.next!=null) { slow = slow.next; fast = fast.next.next; } return slow; } public ListNode sortList(ListNode head) { if(head==null||head.next==null) { return head; } ListNode middle = getMiddleOfList(head); ListNode next = middle.next; middle.next = null; return mergeList(sortList(head), sortList(next)); } //merge the two sorted list ListNode mergeList(ListNode a, ListNode b) { ListNode dummyHead = new ListNode(-1); ListNode curr = dummyHead; while(a!=null&&b!=null) { if(a.val<=b.val) { curr.next=a;a=a.next; } else { curr.next=b;b=b.next; } curr = curr.next; } curr.next = a!=null?a:b; return dummyHead.next; } }
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