leetcode题解（五十一）：148. Sort List

单链表排序
用的是归并排序的思路
递归下去就好

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4
Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

public class Solution {
  
  public ListNode sortList(ListNode head) {
  	//边界条件
    if (head == null || head.next == null)
      return head;
        
    // step 1. cut the list to two halves
    //首先把链表分成两段
    ListNode prev = null, slow = head, fast = head;
    
    while (fast != null && fast.next != null) {
      prev = slow;
      slow = slow.next;
      fast = fast.next.next;
    }
    //慢指针的下一个节点设置为null，九拆分开了
    prev.next = null;
    
    // step 2. sort each half
    //递归的这样走下去
    ListNode l1 = sortList(head);
    ListNode l2 = sortList(slow);
    
    // step 3. merge l1 and l2
    //开始合并链表
    return merge(l1, l2);
  }
  
  ListNode merge(ListNode l1, ListNode l2) {
  	//	初始化
    ListNode l = new ListNode(0), p = l;
    
    while (l1 != null && l2 != null) {
      if (l1.val < l2.val) {
        p.next = l1;
        l1 = l1.next;
      } else {
        p.next = l2;
        l2 = l2.next;
      }
      p = p.next;
    }
    
    if (l1 != null)
      p.next = l1;
    
    if (l2 != null)
      p.next = l2;
    
    return l.next;
  }

}

标签：Sort,head,ListNode,题解,next,五十一,l1,l2,null
来源： https://blog.csdn.net/weixin_43869024/article/details/89737481