Leetcode: Insertion Sort List

Sort a linked list using insertion sort.

 我原本的想法是用额外的空间拷贝每一个节点，建立了一个新的sorted的LinkedList, 后来看到别人的做法不用建立新的LinkedList,  直接以原有List上的节点组成新的sorted的LinkedList。我之前想这样做会扰乱遍历的顺序，但是其实sorted的list和unsorted list是完全分开互不影响的。先还是给sorted list建立一个dummy node做前置节点， 每次取unsorted list里面的一个节点，记录下下一跳位置，然后把这个点插入到sorted list对应位置上（pre.next指到null或者pre.next。val更大）。

Insert Sort的做法相当于是每次从原来的List里面删除头节点，再把这个头节点插入到新的List里相应的位置。这个新List全由原来的节点组成，只是变换了顺序。

时间复杂度是插入排序算法的O(n^2)，空间复杂度是O(1)。

第二遍做法：

public class Solution {
    public ListNode insertionSortList(ListNode head) {
        ListNode dummy = new ListNode(-1);
        ListNode cursor = dummy;
        while (head != null) {
            ListNode next = head.next;
            while (cursor.next!=null && cursor.next.val<=head.val) {
                cursor = cursor.next;
            }
            head.next = cursor.next;
            cursor.next = head;
            head = next;
            cursor = dummy;
        }
        return dummy.next;
    }
}

第一遍做法：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
        public ListNode insertionSortList(ListNode head) {
        if(head == null)
            return null;
        ListNode helper = new ListNode(0);
        ListNode pre = helper;
        ListNode cur = head;
        while(cur!=null)
        {
            ListNode next = cur.next;
            pre = helper;
            while(pre.next!=null && pre.next.val<=cur.val)
            {
                pre = pre.next;
            }
            cur.next = pre.next;
            pre.next = cur;
            cur = next;
        }
        return helper.next;
      }
}

上面程序注释如下：

public ListNode insertionSortList(ListNode head) {
    if(head == null)
        return null;
    ListNode helper = new ListNode(0);//helper is the dummy head for the result sorted LinkedList
    ListNode pre = helper; //pre is the pointer to find the suitable place to plug in the current node
    ListNode cur = head; //cur is current node, the node to be plugged in the sorted list
    while(cur!=null)
    {
        ListNode next = cur.next; //keep a record of the current node's next
        pre = helper;  //after one search, put pre back to its original place
        while(pre.next!=null && pre.next.val<=cur.val) //use pre to traverse the sorted list to find the suitable place to plug in
        {
            pre = pre.next;
        }
        cur.next = pre.next;// plug in current node to the right place
        pre.next = cur;
        cur = next; //go on to deal with the next node in the unsorted list
    }
    return helper.next;
}