List Sort Series

发表于

Insertion Sort List

Question

Sort a linked list using insertion sort.

Analysis

An easy and clear way to sort ( O(1) space )

Code

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode insertionSortList(ListNode head) {
        if(head==null||head.next==null)     return head;
        ListNode next=new ListNode(1);
        ListNode cur=head;
        ListNode res=new ListNode(1);
        ListNode pre=res;
        while(cur!=null){
            next=cur.next;
            while(pre!=null&&pre.next!=null&&pre.next.val<cur.val){
                pre=pre.next;
            }
            cur.next=pre.next;
            pre.next=cur;
            cur=next;
            pre=res;
        }
        
        return res.next;
    }
}

Sort List

Question

Sort a linked list in O(n log n) time using constant space complexity.

Analysis

Java Merge Sort Solution

Code

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if(head==null||head.next==null)     return head;
        
        ListNode slow=head;
        ListNode fast=head;
        ListNode prev=null;
        
        while(fast!=null&&fast.next!=null){
            prev=slow;
            slow=slow.next;
            fast=fast.next.next;
        }
        
        prev.next=null;
        
        ListNode l1=sortList(head);
        ListNode l2=sortList(slow);
        
        return Merge(l1,l2);
    }
    
    public ListNode Merge(ListNode l1, ListNode l2){
        ListNode p=new ListNode(1);
        ListNode res=p;
        
        while(l1!=null&&l2!=null){
            if(l1.val>l2.val){
                res.next=l2;
                l2=l2.next;
            }
            else{
                res.next=l1;
                l1=l1.next;
            }
            res=res.next;
        }
        
        if(l1!=null)
            res.next=l1;
        if(l2!=null)
            res.next=l2;
            
        return p.next;
    }
}