README_EN.md

147. Insertion Sort List

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Description

Sort a linked list using insertion sort.

A graphical example of insertion sort. The partial sorted list (black) initially contains only the first element in the list.

With each iteration one element (red) is removed from the input data and inserted in-place into the sorted list

 

Algorithm of Insertion Sort:

1. Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list.
2. At each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, and inserts it there.
3. It repeats until no input elements remain.

Example 1:

Input: 4->2->1->3

Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0

Output: -1->0->3->4->5

Solutions

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def insertionSortList(self, head: ListNode) -> ListNode:
        if head is None or head.next is None:
            return head
        dummy = ListNode(head.val)
        dummy.next = head
        pre, cur = dummy, head
        while cur:
            if pre.val <= cur.val:
                pre, cur = cur, cur.next
                continue
            p = dummy
            while p.next.val <= cur.val:
                p = p.next
            t = cur.next
            cur.next = p.next
            p.next = cur
            pre.next = t
            cur = t
        return dummy.next

Java

/**
 * 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(head.val);
        dummy.next = head;
        ListNode pre = dummy, cur = head;
        while (cur != null) {
            if (pre.val <= cur.val) {
                pre = cur;
                cur = cur.next;
                continue;
            }
            ListNode p = dummy;
            while (p.next.val <= cur.val) {
                p = p.next;
            }
            ListNode t = cur.next;
            cur.next = p.next;
            p.next = cur;
            pre.next = t;
            cur = t;
        }
        return dummy.next;
    }
}

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