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Sort a linked list using insertion sort.
Given 1->3->2->0->null
, return 0->1->2->3->null
.
分析:
利用insertion sort的原理,把list A的一个node插入到另一个list B里面,这里关键是list B的尾部的处理。
1 /** 2 * Definition for ListNode. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int val) { 7 * this.val = val; 8 * this.next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 /** 14 * @param head: The first node of linked list. 15 * @return: The head of linked list. 16 * cnblogs.com/beiyeqingteng/ 17 */ 18 19 public ListNode insertionSortList(ListNode head) { 20 if (head == null || head.next == null) return head; 21 ListNode dummyNode = new ListNode(Integer.MIN_VALUE); 22 23 while(head != null) { 24 ListNode current = dummyNode; 25 while(head.val > current.val && current.next != null && current.next.val < head.val) { 26 current = current.next; 27 } 28 ListNode remainingHead = current.next; 29 current.next = head; 30 head = head.next; 31 current.next.next = remainingHead; 32 } 33 return dummyNode.next; 34 } 35 }
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原文地址:http://www.cnblogs.com/beiyeqingteng/p/5636402.html