Reference: LeetCode
Difficulty: Medium

My Post: [3ms] Java solution with comments (visual)

Problem

Sort a linked list using insertion sort.

Example:

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

Analysis

Iteration

Original: 33 ms

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public ListNode insertionSortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}

ListNode dummy = new ListNode(-1);
dummy.next = head;

ListNode prev = dummy;
ListNode curr = head;
// 1 3 5 7 4 9
// p p.n pr curr nT
while (curr != null) {
ListNode nextTemp = curr.next; // record the next node to proceed
ListNode p = dummy;
while (p.next != curr && p.next.val <= curr.val) {
p = p.next;
} // p.next stops at the node that is > curr.val

if (p.next != curr) { // modify when p.next is not curr itself
prev.next = curr.next; // remove
curr.next = p.next; // add
p.next = curr;
} else {
prev = curr; // if p.next == curr, we need to update prev
}
// update
curr = nextTemp;
}

return dummy.next;
}

Improvement: 3 ms

Note:

  • prev != dummy is very important!!! Otherwise, dummy node would be compared.
  • Or set prev = head and curr = head.next.
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public ListNode insertionSortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}

ListNode dummy = new ListNode(-1);
dummy.next = head;

ListNode prev = dummy;
ListNode curr = head;
// 1 3 5 7 4 9
// p p.n pr curr nT
while (curr != null) {
ListNode nextTemp = curr.next;
ListNode p = dummy;

if (prev != dummy && prev.val > curr.val) {
while (p.next != curr && p.next.val <= curr.val) {
p = p.next;
} // p.next stops at the node that is > curr.val
prev.next = curr.next; // remove
curr.next = p.next; // add
p.next = curr;
// do not need to update prev
} else {
prev = curr; // update prev, since curr is not modified
}
// update
curr = nextTemp;
}

return dummy.next;
}

The version I wrote at the first time:

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// dummy  3    4    1    5
// t t.n p p.n
// prev
public ListNode insertionSortList(ListNode head) {
if (head == null || head.next == null) { // size == 0 OR 1
return head;
}
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode p = head.next; // start from the second node
ListNode prev = head; // otherwise it would compare with dummy

while (p != null) {
ListNode t = dummy;

if (prev.val > p.val) { // p needs swapping (better performance)
while (t.next != p && p.val > t.next.val) {
t = t.next;
} // stops at the right position
prev.next = p.next;
p.next = t.next;
t.next = p;
p = prev.next; // set next (prev not changing)
} else { // p needs not swapping
prev = p;
p = p.next;
}
}
return dummy.next;
}

Time: $O(N^2)$
Space: $O(1)$


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