題目描述
Sort a linked list in O(n log n) time using constant space complexity.
歸併排序算法:時間複雜度是O(logn),對於數組,空間複雜度是O(n),對於鏈表空間複雜度是O(1)
鏈表的歸併排序:
(1)根據快慢指針知道中間的指針
(2)把鏈表從中間位置的指針斷開
(3)遞歸調用,不喝合併兩個有序的兩部分鏈表
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode sortList(ListNode head) {
if(head==null || head.next==null) return head;
ListNode mid=getMid(head);
ListNode midnext=mid.next;
mid.next=null;//斷開
return MergeList(sortList(head),sortList(midnext));
}
private ListNode MergeList(ListNode left, ListNode right) {
if(left==null) return right;
if(right==null) return left;
ListNode head=null;//臨時鏈表保存合併以後的鏈表
if(left.val<right.val){
head=left;
left=left.next;
}else{
head=right;
right=right.next;
}
ListNode temp=head;
while(left!=null && right!=null){
if(left.val<right.val){
temp.next=left;
left=left.next;
}else{
temp.next=right;
right=right.next;
}
temp=temp.next;
}
if(right!=null){
temp.next=right;
}
if(left!=null){
temp.next=left;
}
return head;
}
private ListNode getMid(ListNode head) {
ListNode low=head;
ListNode fast=head;
while(fast.next!=null&&fast.next.next!=null){
low=low.next;
fast=fast.next.next;
}
return low;
}
}
題目描述
Sort a linked list using insertion sort.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode insertionSortList(ListNode head) {
ListNode dumy=new ListNode(Integer.MIN_VALUE);
ListNode pre=dumy;
ListNode cur=head;
while(cur!=null){
ListNode next=cur.next;//保存當前待插入節點的下一個節點
pre=dumy;
while(pre.next!=null&&pre.next.val<cur.val){
pre=pre.next;//前插法,從新的鏈表從前到後找當前節點的位置
}
//循環出來以後找到當前節點的正確位置就是pre目前的位置
//插入當前節點
cur.next=pre.next;
pre.next=cur;
//出來下一個節點
cur=next;
}
return dumy.next;
}
}