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MergeSort_LinkedList.java
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99 lines (78 loc) · 2.74 KB
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package Algorithms.sort;
import java.util.LinkedList;
import Algorithms.algorithm.others.ListNode;
public class MergeSort_LinkedList {
public static void main(String[] args) {
ListNode node1 = new ListNode(0);
ListNode node2 = new ListNode(15);
ListNode node3 = new ListNode(6);
ListNode node4 = new ListNode(9);
ListNode node5 = new ListNode(4);
node1.next = node2;
node2.next = node3;
node3.next = node4;
node4.next = node5;
System.out.println(node1.toString());
sort(node1);
System.out.println(node1.toString());
}
public static ListNode sort(ListNode head) {
if (head == null) {
return null;
}
// 注意一定要加这一行，否则会不断对1个元素无限分下去
if (head.next == null) {
return head;
}
ListNode mid = findMidPre(head);
// 将list切为2个list.
ListNode right = mid.next;
mid.next = null;
//调用将2边分别排序
ListNode left = sort(head);
right = sort(right);
// 将2个已经排序的List Merge在一起
return merge(left, right);
}
public static ListNode merge(ListNode head1, ListNode head2) {
ListNode dummy = new ListNode(0);
// cur 表示新链的尾部.
ListNode cur = dummy;
while (head1 != null && head2 != null) {
// 将2个链中较小的一个接到新链的尾部
if (head1.val < head2.val) {
cur.next = head1;
head1 = head1.next;
} else {
cur.next = head2;
head2 = head2.next;
}
// 将扫描指针移动到新链的尾部0
cur = cur.next;
}
// 把未扫描完毕的链接在新链的结尾即可
if (head1 != null) {
cur.next = head1;
} else {
cur.next = head2;
}
// 返回新链的头部
return dummy.next;
}
// 这个函数是寻找Mid的前一个节点，
// 技巧就是：一开始就将Fast放在head的前一个节点，这样当只有2个节点的时候：
// 1->2->null
// slow 会停在1处 ，这样就可以处理只有2个节点的情况.
public static ListNode findMidPre(ListNode head) {
if (head == null) {
return null;
}
ListNode fast = head.next;
ListNode slow = head;
if (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
}
return slow;
}
}