LeetCode Solution – Sort a linked list using insertion sort in Java
Insertion Sort List:
Sort a linked list using insertion sort.
This is my accepted answer for LeetCode problem - Sort a linked list using insertion sort in Java. It is a complete program.
Before coding for that, here is an example of insertion sort from wiki. You can get an idea of what is insertion sort.
Code:
package algorithm.sort; class ListNode { int val; ListNode next; ListNode(int x) { val = x; next = null; } } public class SortLinkedList { public static ListNode insertionSortList(ListNode head) { if (head == null || head.next == null) return head; ListNode newHead = new ListNode(head.val); ListNode pointer = head.next; // loop through each element in the list while (pointer != null) { // insert this element to the new list ListNode innerPointer = newHead; ListNode next = pointer.next; if (pointer.val <= newHead.val) { ListNode oldHead = newHead; newHead = pointer; newHead.next = oldHead; } else { while (innerPointer.next != null) { if (pointer.val > innerPointer.val && pointer.val <= innerPointer.next.val) { ListNode oldNext = innerPointer.next; innerPointer.next = pointer; pointer.next = oldNext; } innerPointer = innerPointer.next; } if (innerPointer.next == null && pointer.val > innerPointer.val) { innerPointer.next = pointer; pointer.next = null; } } // finally pointer = next; } return newHead; } public static void main(String[] args) { ListNode n1 = new ListNode(2); ListNode n2 = new ListNode(3); ListNode n3 = new ListNode(4); ListNode n4 = new ListNode(3); ListNode n5 = new ListNode(4); ListNode n6 = new ListNode(5); n1.next = n2; n2.next = n3; n3.next = n4; n4.next = n5; n5.next = n6; n1 = insertionSortList(n1); printList(n1); } public static void printList(ListNode x) { if(x != null){ System.out.print(x.val + " "); while (x.next != null) { System.out.print(x.next.val + " "); x = x.next; } System.out.println(); } } } |
Output:
2 3 3 4 4 5
<pre><code> String foo = "bar"; </code></pre>
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gao can
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Rupam Ranjan Rana
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junmin
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Zack