题目描述
给定一个无序单链表，实现单链表的排序(按升序排序)。
示例
输入：head = [4,2,1,3]
输出：[1,2,3,4]
解决方法一：
时间复杂度：O(nlogn)
空间复杂度：O(logn)
归并排序(递归)
- 首先用快慢指针找到链表的中点
- 对链表进行分割
- 对链表进行合并(合并两个有序链表)
import java.util.*;
/*
* public class ListNode {
* int val;
* ListNode next = null;
* }
*/
public class Solution {
/**
*
* @param head ListNode类 the head node
* @return ListNode类
*/
public ListNode sortInList (ListNode head) {
/*
1. 用快慢指针找到链表中点
2. 分割链表
3. 借助两个链表合并思想，对链表进行合并
*/
if (head == null || head.next == null) {
return head;
}
ListNode slow = head;
ListNode fast = head.next;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
ListNode rightNode = slow.next;
slow.next = null;
ListNode left = sortInList(head); //分割前半段
ListNode right = sortInList(rightNode); //分割后半段
return merge(left, right); //合并两个有序链表
}
public ListNode merge (ListNode phead1, ListNode phead2) {
if (phead1 == null || phead2 == null) {
return phead1 == null ? phead2 : phead1;
}
ListNode phead = new ListNode(-1);
ListNode ptail = phead;
ListNode l1 = phead1, l2 = phead2;
while (l1 != null && l2 != null) {
if (l1.val >= l2.val) {
ptail.next = l2;
l2 = l2.next;
} else {
ptail.next = l1;
l1 = l1.next;
}
ptail = ptail.next;
}
ptail.next = (l1 == null ? l2 : l1);
return phead.next;
}
}
解决方法二：
时间复杂度：O(nlogn)
空间复杂度：O(1)
使用自底向上的方法实现归并排序，则可以达到 O(1)的空间复杂度。
首先求得链表的长度length，然后将链表拆分成子链表进行合并。
import java.util.*;
/*
* public class ListNode {
* int val;
* ListNode next = null;
* }
*/
public class Solution {
/**
*
* @param head ListNode类 the head node
* @return ListNode类
*/
public ListNode sortInList (ListNode head) {
/*
1. 用快慢指针找到链表中点
2. 分割链表
3. 借助两个链表合并思想，对链表进行合并
*/
//==>递归：空间复杂度不是O(1)
//==>迭代
if (head == null || head.next == null) {
return head;
}
int len = 0;
ListNode temp = head;
while (temp != null) {
len++;
temp = temp.next;
}
ListNode phead = new ListNode(0);
phead.next = head;
for (int subLen = 1; subLen < len; subLen <<= 1) {
//1 -> 2 -> 4 -> 8 ...
ListNode preNode = phead, ptail = phead.next;
while (ptail != null) {
ListNode phead1 = ptail;
for (int i = 1; i < subLen && ptail.next != null; i++) {
ptail = ptail.next;
}
ListNode phead2 = ptail.next;
ptail.next = null;
ptail = phead2;
for (int i = 1; i < subLen && ptail != null && ptail.next != null; i++) {
ptail = ptail.next;
}
ListNode flag = null; //记录还未处理的后大段链表
if (ptail != null) {
flag = ptail.next;
ptail.next = null;
}
preNode.next = merge(phead1, phead2);
while (preNode.next != null) {
preNode = preNode.next;
}
ptail = flag;
}
}
return phead.next;
}
public ListNode merge (ListNode phead1, ListNode phead2) {
if (phead1 == null || phead2 == null) {
return phead1 == null ? phead2 : phead1;
}
ListNode phead = new ListNode(-1);
ListNode ptail = phead;
ListNode l1 = phead1, l2 = phead2;
while (l1 != null && l2 != null) {
if (l1.val >= l2.val) {
ptail.next = l2;
l2 = l2.next;
} else {
ptail.next = l1;
l1 = l1.next;
}
ptail = ptail.next;
}
ptail.next = (l1 == null ? l2 : l1);
return phead.next;
}
}