[LeetCode]Insertion Sort List

Sort a linked list using insertion sort.

Analysis

The
general idea is insert the current element A[i] into the proper position from A[0]...A[i-1], and A[0]...A[i-1] is already sorted.

In this problem, we can use the same idea and linked list provides a more efficient way for insertion. Details can be
found in the code below. Note that after the insertion, the position of P is unchanged but should not provide another p=p->next operation.

Complexity
is O(n^2)

java

public ListNode insertionSortList(ListNode head) {
        if(head == null || head.next == null) return head;
        ListNode newHead = new ListNode(-1);
        newHead.next = head;
        ListNode cur = head;
        ListNode post = head.next;
        while(post!=null){
        	if(post.val>=cur.val){
        		cur = cur.next;
        		post = post.next;
        	}else {
				ListNode insertCur = newHead;
				ListNode insertPost = newHead.next;
				while(insertPost.val<post.val){
					insertCur = insertPost;
					insertPost = insertPost.next;
				}
				cur.next = post.next;
				post.next = insertPost;
				insertCur.next = post;
				post = cur.next;
			}
        }
        return newHead.next;
    }

c++

ListNode *insertionSortList(ListNode *head) {
        if(head == NULL) return head;
        ListNode *p = new ListNode(-1);
        p->next = head;
        ListNode *pre = head;
        ListNode *cur = pre->next;
        while(cur){
            
            if(cur->val >= pre->val){
                pre = cur;
                cur = cur->next;
            }else{
                ListNode *insertPre = p;
                ListNode *insertCur = p->next;
                while(insertCur->val < cur->val){
                    insertPre = insertCur;
                    insertCur = insertCur->next;
                }
                pre->next = cur->next;
                cur->next = insertCur;
                insertPre->next = cur;
                cur = pre->next;
            }
            
        }
        head = p->next;
        return head;
    }