# Insertion Sort List LeetCode Solution | Easy Approach Share:

Insertion Sort Given the `head` of a singly linked list, sort the list using insertion sort, and return the sorted list’s head.

The steps of the insertion sort algorithm:

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.

The following is a graphical example of the insertion sort algorithm. The partially sorted list (black) initially contains only the first element in the list. One element (red) is removed from the input data and inserted in-place into the sorted list with each iteration.

Example 1:

```Input: head = [4,2,1,3]
Output: [1,2,3,4]
```

Example 2:

```Input: head = [-1,5,3,4,0]
Output: [-1,0,3,4,5]
```

Constraints:

• The number of nodes in the list is in the range `[1, 5000]`.
• `-5000 <= Node.val <= 5000`

Time: O(n)
Space: O(1)

### C++

`````` class Solution {
public:
ListNode dummy(0);
ListNode* prev = &dummy;  // the last (largest) of the sorted list

while (head) {                  // current inserting node
ListNode* next = head->next;  // cache next inserting node
if (prev->val >= head->val)   // `prev` >= current inserting node
prev = &dummy;              // move `prev` to the front
while (prev->next && prev->next->val < head->val)
prev = prev->next;
head = next;  // update current inserting node
}

return dummy.next;
}
};
``````

### Java

`````` class Solution {
ListNode dummy = new ListNode(0);
ListNode prev = dummy; // the last (largest) of the sorted list

while (head != null) {       // current inserting node
ListNode next = head.next; // cache next inserting node
if (prev.val >= head.val)  // `prev` >= current inserting node
prev = dummy;            // move `prev` to the front
while (prev.next != null && prev.next.val < head.val)
prev = prev.next;
head = next; // update current inserting node
}

return dummy.next;
}
}
``````

### Python

``````class Solution:
def insertionSortList(self, head: ListNode) -> ListNode:
dummy = ListNode(0)

while curr:
prev = dummy
while prev.next and prev.next.val < curr.val:
prev = prev.next
next = curr.next
curr.next = prev.next
prev.next = curr
curr = next

return dummy.next

``````

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