Sort List

Sort a linked list in O(n log n) time using constant space complexity.

 

Merge sort

 先根据mid node把list 分割为 单个 node ，然后merge

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
       if(head == null || head.next == null)
        return head;
        ListNode fast = head.next.next;
        ListNode p = head;
        //  while结束时 p指向的是中点的前一个节点
        while(fast != null && fast.next != null){
            p = p.next;
            fast = fast.next.next;
        }
        // p is the mid node of the list // 这两行代码顺序不能换
        ListNode h2 = sortList(p.next);
        p.next = null;
        // 把前半部分的merge sort结果 和后半部分的merge sort结果再merge sort
        return mergeSort(sortList(head), h2);
    }
    
    // merge two sorted list
    private ListNode mergeSort(ListNode h1, ListNode h2){
        ListNode fakeNode = new ListNode(Integer.MIN_VALUE);
        ListNode runner = fakeNode;
        while(h1 != null && h2 != null){
            if(h1.val < h2.val){
                runner.next = h1;
                h1 = h1.next;
            }else{
                runner.next = h2;
                h2 = h2.next;
            }
            runner = runner.next;
        }
        if(h1 != null){
            runner.next = h1;
        }
        if(h2 != null){
            runner.next = h2;
        }
        return fakeNode.next;
    }
}