标签:style blog io color sp for on div log
Sort a linked list using insertion sort.
这道题用两个链表做的,这样方便一点。还有新链表头结点我没有存放内容,这样也比较方便,后面返回head1.next就好了
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public ListNode insertionSortList(ListNode head) { 14 ListNode head1 = new ListNode(0); 15 if(null == head || null == head.next) 16 return head; 17 18 else{ 19 ListNode temp = new ListNode(0); 20 temp.val = head.val; 21 temp.next = null; 22 head1.next = temp;//用一个新链表 23 24 ListNode ptr = head.next; 25 while(ptr != null){ //遍历原来的链表,插入到新链表中 26 temp = head1.next; 27 ListNode tempPre = head1; //temp前面的节点 28 while(temp != null){ 29 if(temp.val > ptr.val){ //在temp前面插入结点 30 ListNode nodeAdd = new ListNode(0); 31 nodeAdd.val = ptr.val; 32 nodeAdd.next = temp; 33 tempPre.next = nodeAdd; 34 break; //插入完成,跳出循环 35 } 36 temp = temp.next; 37 tempPre = tempPre.next; 38 } 39 if(null == temp){ //在新链表添加节点 40 ListNode nodeAdd = new ListNode(0); 41 nodeAdd.val = ptr.val; 42 nodeAdd.next = null; 43 tempPre.next = nodeAdd; 44 } 45 ptr = ptr.next; 46 } 47 } 48 49 return head1.next; 50 } 51 }
标签:style blog io color sp for on div log
原文地址:http://www.cnblogs.com/luckygxf/p/4119865.html