LeetCode Sort List 解题报告

Sort a linked list in O(n log n) time using constant space complexity.
http://oj.leetcode.com/problems/sort-list/

解题报告：就是对一个链表进行归并排序。
主要考察3个知识点，
知识点1：归并排序的整体思想
知识点2：找到一个链表的中间节点的方法
知识点3：合并两个已排好序的链表为一个新的有序链表

归并排序的基本思想是：找到链表的middle节点，然后递归对前半部分和后半部分分别进行归并排序，最后对两个以排好序的链表进行Merge。

AC代码： https://github.com/tanglu/Leetcode/blob/master/sortList.java

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
     //use the fast and slow pointer to get the middle of a ListNode
     ListNode getMiddleOfList(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        while(fast.next!=null&&fast.next.next!=null) {
        	slow = slow.next;
        	fast = fast.next.next;
        }
        return slow;
    }
    
    public ListNode sortList(ListNode head) {
        if(head==null||head.next==null) {
            return head;
        }
    	ListNode middle = getMiddleOfList(head);
    	ListNode next   = middle.next;
    	    middle.next = null;
    	return mergeList(sortList(head), sortList(next));
    }
    
    //merge the two sorted list
    ListNode mergeList(ListNode a, ListNode b) {
    	ListNode dummyHead = new ListNode(-1);
    	ListNode curr = dummyHead;
    	while(a!=null&&b!=null) {
    		if(a.val<=b.val) {
    			curr.next=a;a=a.next;
    		} else {
    			curr.next=b;b=b.next;
    		}
    		curr  = curr.next;
    	}
    	curr.next = a!=null?a:b;
    	return dummyHead.next;
    }
}