PROBLEM DESCRIPTION
Given the head of a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. You should preserve the original relative order of the nodes in each of the two partitions.
SOLUTION
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/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode partition(ListNode head, int x) {
// Create two new linked lists to hold nodes less than x and nodes greater than or equal to x.
ListNode lessHead = new ListNode(101); // Placeholder value, will be ignored.
ListNode lessTail = lessHead; // Pointer to the tail of the "less than x" list.
ListNode moreHead = new ListNode(101); // Placeholder value, will be ignored.
ListNode moreTail = moreHead; // Pointer to the tail of the "greater than or equal to x" list.
ListNode t = head; // Temporary pointer to traverse the original linked list.
while(t != null){
ListNode next = t.next; // Save the next node before modifying current node's next pointer.
// If the current node's value is less than x, append it to the "less than x" list.
if(t.val < x){
lessTail.next = t; // Add current node to the end of the "less than x" list.
t.next = null; // Break the next pointer of the current node.
lessTail = t; // Move the tail pointer of the "less than x" list to the current node.
// If the current node's value is greater than or equal to x, append it to the "greater than or equal to x" list.
} else {
moreTail.next = t; // Add current node to the end of the "greater than or equal to x" list.
t.next = null; // Break the next pointer of the current node.
moreTail = t; // Move the tail pointer of the "greater than or equal to x" list to the current node.
}
t = next; // Move to the next node in the original list.
}
// Connect the tail of the "less than x" list to the head of the "greater than or equal to x" list. (skipping the dummy note)
lessTail.next = moreHead.next;
// The "less than x" list now contains all nodes less than x in their original order,
// followed by all nodes greater than or equal to x in their original order.
return lessHead.next; // Return the head of the merged list.
}
}