标签：style blog color io 使用 for div sp log
Sort a linked list using insertion sort.
难度：84. 我自己的做法是使用了额外的空间，建立了一个新的sorted的LinkedList, 技巧还是建立一个dummy node做前置节点。
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 if(head == null) 15 return null; 16 ListNode helper = new ListNode(0); 17 ListNode pre = helper; 18 ListNode cur = head; 19 while(cur!=null) 20 { 21 ListNode next = cur.next; 22 pre = helper; 23 while(pre.next!=null && pre.next.val<=cur.val) 24 { 25 pre = pre.next; 26 } 27 cur.next = pre.next; 28 pre.next = cur; 29 cur = next; 30 } 31 return helper.next; 32 } 33 }
然后网上看到别人不用建立一个新的LinkedList, 直接在原来的LinkedList上面完成sort（我之前想过，但是一想到这样会有可能把遍历顺序搞乱，所以作罢，想到建立一个新的LinkedList），这个人记录了当前节点的下一跳的位置，然后把当前节点加入到已经sorted的list里面，不会影响。好赞的做法
难度：93
1 public ListNode insertionSortList(ListNode head) { 2 if(head == null) 3 return null; 4 ListNode helper = new ListNode(0);//helper is the dummy head for the result sorted LinkedList 5 ListNode pre = helper; //pre is the pointer to find the suitable place to plug in the current node 6 ListNode cur = head; //cur is current node, the node to be plugged in the sorted list 7 while(cur!=null) 8 { 9 ListNode next = cur.next; //keep a record of the current node‘s next 10 pre = helper; 11 while(pre.next!=null && pre.next.val<=cur.val) //use pre to traverse the sorted list to find the suitable place to plug in 12 { 13 pre = pre.next; 14 } 15 cur.next = pre.next;// plug in current node to the right place 16 pre.next = cur; 17 cur = next; //go on to deal with the next node in the unsorted list 18 } 19 return helper.next; 20 }
标签：style blog color io 使用 for div sp log
原文地址：http://www.cnblogs.com/EdwardLiu/p/3976821.html
