leetcode148-Sort List

题目

给你链表的头结点 head ，请将其按 升序 排列并返回 排序后的链表 。
示例 1：
输入：head = [4,2,1,3]
输出：[1,2,3,4]

分析

这道题目的思路也比较明确，我们可以用递归排序的思路，先不断拆分链表直到只剩下一个结点，然后再合并俩个有序列表。拆分链表可以用快慢指针的方法

public class LinkNode {
	int val;
	LinkNode next;

	public LinkNode(int data) {
		this.val = data;
		this.next = null;
	}
}
public class LinkList {
	LinkNode head;
	public LinkList() {
		this.head = null;
	}
	public LinkNode getHead() {
		return this.head;
	}
	//添加元素
	public void addNode(int data) {
		LinkNode node = new LinkNode(data);
		if (this.head == null) {
			this.head = node;
		} else {
			LinkNode cur = this.head;
			while(cur.next != null) {
				cur = cur.next;
			}
			cur.next = node;
		}
	}
	//正序打印
	public void print(LinkNode node) {
		while(node != null) {
			System.out.print(node.val);
			System.out.print(" ");
			node = node.next;
		}
		System.out.println();
	}
	public LinkNode sort(LinkNode head) {
		if(head == null || head.next == null) {
			return head;
		}
		LinkNode fast = head;
		LinkNode slow = head;
		LinkNode pre = null;
		while(fast != null && fast.next != null) {
			pre = slow;
			slow = slow.next;
			fast = fast.next.next;
		}
		pre.next = null;
		return merge(sort(slow),sort(head));
	}
	public LinkNode merge(LinkNode slow,LinkNode fast) {
		LinkNode dummy = new LinkNode(0);
		LinkNode p = dummy;
		while(slow != null && fast != null) {
			if(slow.val < fast.val) {
				p.next = slow;
				p = p.next;
				slow = slow.next;
			} else {
				p.next = fast;
				p = p.next;
				fast = fast.next;
			}
		}
		while(slow != null) {
			p.next = slow;
			p = p.next;
			slow = slow.next;
		}
		while(fast != null) {
			p.next = fast;
			p = p.next;
			fast = fast.next;
		}
		return dummy.next;
	}
}
public class sortList {
	public static void main(String[] args) {
		LinkList list = new LinkList();
		list.addNode(-1);
		list.addNode(5);
		list.addNode(3);
		list.addNode(4);
		list.addNode(0);
		LinkNode head = list.sort(list.getHead());
		list.print(head);
	}
}
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