算法-插入排序链表

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
主要思路： 递归
先递归到最后一个节点，然后在每次向上回溯时进行排序
 */
class Solution {
    public ListNode insertionSortList(ListNode head) {
        if(head == null || head.next == null){
            return head;
        }
        ListNode tmpHead = insertionSortList(head.next);
        if(head.val <= tmpHead.val){ // 如果两个节点相等，则插入到已存在节点的前面
            head.next = tmpHead;
            return head;
        }else{
            ListNode tmpHead2 = tmpHead;
            // 找到合适的位置的前一个节点
            while(tmpHead2.next != null && tmpHead2.next.val < head.val){
                tmpHead2 = tmpHead2.next;
            }
            // 将head连接到链表中
            ListNode b = tmpHead2.next;
            tmpHead2.next = head;
            head.next = b;
            return tmpHead;            
        }
    }
}


// 头插法
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode insertionSortList(ListNode head) {
        if(head == null || head.next == null){
            return head;
        }
        ListNode dummy = new ListNode();
        dummy.next = head;

        ListNode cur = head;
        while(cur.next != null){
            ListNode newcur = dummy;
            while(true){
                if(newcur.next == cur.next){
                    cur = cur.next;
                    break;
                }
                if(newcur.next.val >= cur.next.val){
                    ListNode curNext = cur.next;
                    // 先给cur.next节点从链表中提取出来
                    cur.next = cur.next.next;
                    // 将cur.next连接到newcur.next上
                    ListNode tmpNext = newcur.next;
                    newcur.next = curNext;
                    newcur.next.next = tmpNext;
                    break;
                }else{
                    newcur = newcur.next;
                }
            }
        }
        return dummy.next;
    }
}