147. Insertion Sort List

题目：

Sort a linked list using insertion sort. 

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 Linked List Sort 

链接： https://leetcode.com/problems/insertion-sort-list/

6/12/2017

46ms, 17%

从head开始遍历，碰到相同或者大的就插入，更新next指针

注意

1. 需要用sortedLength来区分已经排好序的部分，或许不用？

2. 设一个dummy node这样prev部分比较好解决

3. 排好序的部分最后一个元素next = null，这样才能在插入之后不形成环

4. 第27行判断是否刚加入的node是排序部分最大的，是的话将其next = null

5. 外循环时每次先保存next，不然内循环会被覆盖。第14行

public class Solution {
    public ListNode insertionSortList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        int sortedLength = 1;
        ListNode cur = head.next;
        ListNode next;
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        head.next = null;

        while (cur != null) {
            next = cur.next;
            ListNode prev = dummy, sorted = dummy.next;
            int i = 0;
            for (; i < sortedLength; i++) {
                if (sorted.val >= cur.val) {
                    prev.next = cur;
                    cur.next = sorted;
                    break;
                } else {
                    prev = sorted;
                    sorted = sorted.next;
                }
            }
            if (i == sortedLength) {
                prev.next = cur;
                cur.next = null;
            }
            sortedLength++;
            cur = next;
        }
        return dummy.next;
    }
}

别人的算法精炼很多，看来sortedLength不需要。做法类似于reverse linkedlist, dummy.next最开始是null，所以之前第3点可以保证。而且可以把所有插入的case统一到一种形式，比我的break和i == sortedLength要好很多

public ListNode insertionSortList(ListNode head) {
        if( head == null ){
            return head;
        }
        
        ListNode helper = new ListNode(0); //new starter of the sorted list
        ListNode cur = head; //the node will be inserted
        ListNode pre = helper; //insert node between pre and pre.next
        ListNode next = null; //the next node will be inserted
        //not the end of input list
        while( cur != null ){
            next = cur.next;
            //find the right place to insert
            while( pre.next != null && pre.next.val < cur.val ){
                pre = pre.next;
            }
            //insert between pre and pre.next
            cur.next = pre.next;
            pre.next = cur;
            pre = helper;
            cur = next;
        }
        
        return helper.next;
    }

更多讨论

https://discuss.leetcode.com/category/155/insertion-sort-list