标签：完整 递归 blog should list 还需要 com complex code
Sort a linked list in O(n log n) time using constant space complexity.
Example
Given 1->3->2->null, sort it to 1->2->3->null.
思路：如果空间不限制，完全可以把所有的val，拿出来排序之后，在从头到尾设置val，不用动指针
本题思路和一般的排序一样，用快速排序或者归并排序
归并排序思路：
需要写一个函数，将两个有序链表排序。这个可以使用最简单的方法非递归的
/** * Definition for ListNode. * public class ListNode { * int val; * ListNode next; * ListNode(int val) { * this.val = val; * this.next = null; * } * } */ public class Solution { /** * @param ListNode l1 is the head of the linked list * @param ListNode l2 is the head of the linked list * @return: ListNode head of linked list */ public ListNode mergeTwoLists(ListNode l1, ListNode l2) { // write your code here ListNode head1 = l1; ListNode head2 = l2; ListNode ans=new ListNode(0); ListNode anss=ans; while(head1!=null&&head2!=null){ if(head1.val<head2.val){ ListNode temp=head1.next; head1.next=null; ans.next=head1; head1=temp; }else{ ListNode temp=head2.next; head2.next=null; ans.next=head2; head2=temp; } ans=ans.next; } while(head1!=null){ ListNode temp=head1.next; head1.next=null; ans.next=head1; head1=temp; ans=ans.next; } while(head2!=null){ ListNode temp=head2.next; head2.next=null; ans.next=head2; head2=temp; ans=ans.next; } return anss.next; } }
递归版本的：
public ListNode mergeTwoLists(ListNode l1, ListNode l2) { // write your code here if(l1==null){ return l2; } if(l2==null){ return l1; } ListNode head; if(l1.val<l2.val){ head=l1; head.next=mergeTwoLists(l1.next,l2); }else{ head=l2; head.next=mergeTwoLists(l1,l2.next); } return head; }
这两种各有特点。
还需要用一个快慢指针的方法找中点。
剩下的就是归并的思路。
上完整代码
/** * Definition for ListNode. * public class ListNode { * int val; * ListNode next; * ListNode(int val) { * this.val = val; * this.next = null; * } * } */ public class Solution { /** * @param head: The head of linked list. * @return: You should return the head of the sorted linked list, using constant space complexity. */ public ListNode sortList(ListNode head) { // write your code here ListNode ans=sort(head); return ans; } public ListNode sort(ListNode head){ if(head==null||head.next==null){ return head; } ListNode slow=head; ListNode fast=head.next; while(fast!=null&&fast.next!=null){ fast=fast.next.next; slow=slow.next; } ListNode head2=slow.next; slow.next=null; ListNode p1=sort(head); ListNode p2=sort(head2); ListNode p = mergeTwoLists(p1,p2); return p; } public ListNode mergeTwoLists(ListNode l1, ListNode l2) { // write your code here if(l1==null){ return l2; } if(l2==null){ return l1; } ListNode head; if(l1.val<l2.val){ head=l1; head.next=mergeTwoLists(l1.next,l2); }else{ head=l2; head.next=mergeTwoLists(l1,l2.next); } return head; } }
标签：完整 递归 blog should list 还需要 com complex code
原文地址：http://www.cnblogs.com/tobemaster/p/6034649.html
