[LeetCode刷题] Insertion Sort List

这个是刷的第一题，medium难度。算法没什么难度，主要是有一个edge case。maintain两个list，第一个是sort好的，第二个是没sort得部分。值得重新做的一个题目，有些小技巧在下面另一个解法讲。

/**
 * 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 ret = head;
        ListNode cur = head.next;
        ret.next = null;
        while(cur != null) {
            ListNode rCur = ret;
            boolean found = false;
            while(rCur.next != null) {
                if(cur.val <= rCur.val) {
                    //Insert to head
                    ListNode temp = cur.next;
                    cur.next = rCur;
                    ret = cur;
                    cur = temp;
                    found = true;
                    break;
                }
                if(cur.val >= rCur.val && cur.val <= rCur.next.val) {
                    //insert to middle
                    ListNode temp = cur.next;
                    cur.next = rCur.next;
                    rCur.next = cur;
                    cur = temp;
                    found = true;
                    break;
                }
                rCur = rCur.next;
            }
            if(!found) {
                if(cur.val <= rCur.val) {//edge case, 如果出现这样的情况，说明这整个list只有两个元素。
                    ListNode temp = cur.next;
                    cur.next = rCur;
                    ret = cur;
                    cur = temp;
                } else {
                //insert to tail
                    ListNode temp = cur.next;
                    cur.next = null;
                    rCur.next = cur;
                    cur = temp;
                }
            }
        }
        return ret;
    }
}

看到了一个别人更好的解法，使用了一个helper在head之前，这样少了那个edge case，简便了很多

public class Solution {
    public ListNode insertionSortList(ListNode head) {
        if(head == null) {
            return null;
        }
        //这里用一个技巧，省去了head的edge case
        ListNode helper = new ListNode(0);
        //这样cur直接定在head即可
        ListNode cur = head;
        ListNode pre;
        while(cur != null) {
            ListNode next = cur.next;
            pre = helper;
            //注意这里的比较，找insert的position时，比较只需要比后面一个与cur的大小
            while(pre.next != null && pre.next.val <= cur.val) {
                pre = pre.next;
            }
            cur.next = pre.next;
            pre.next = cur;
            cur = next;
        }
        return helper.next;


    }
}