LeetCode#147. Insertion Sort List

• 题目：利用插入排序对链表进行排序
• 难度：Medium
• 思路：思路同数组的插入排序（第i趟排序保证第i+1个元素处于前i个元素中的正确位置上），但是链表的指针变化需要注意（在这个题目中需要定义4个指针才能调整成功）
• 代码：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode insertionSortList(ListNode head) {        if(head == null || head.next == null){            return head;        }        int len = 0;        ListNode node = head;        while(node != null){            len++;            node = node.next;        }        ListNode newHead = new ListNode(0);        newHead.next = head;        for(int i=1; i < len; i++){            int step = i;            ListNode compareNode = newHead;//第i趟插入，第i+1个元素找到自己的位置            ListNode  beforeCompare = compareNode;            while(step >= 0){                beforeCompare = compareNode;                compareNode = compareNode.next;                step--;            }            ListNode pre = newHead;            ListNode next = pre.next;            int j = 1;            while(j <= i && next.val <= compareNode.val){                pre = next;                next = next.next;                j++;            }            //如果前面没有元素比compareNode的值大，就不需要修改指针           if(pre != beforeCompare){                pre.next = compareNode;                beforeCompare.next = compareNode.next;                compareNode.next = next;            }        }        return newHead.next;    }}

• Discuss里简洁的算法（只用了3个临时指针）

public class Solution {    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;    }}