LeetCode 86: Partition List Solution

Master LeetCode problem 86 (Partition List), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

86. Partition List

Medium

Linked List Two Pointers

Problem Explanation

Explanation

To partition the linked list such that all nodes less than x come before nodes greater than or equal to x, we can maintain two separate linked lists - one for nodes less than x and the other for nodes greater than or equal to x. Then, we can merge these two lists at the end. We iterate through the original list and append nodes to their respective lists based on the comparison with x. Finally, we connect the tail of the "less than x" list to the head of the "greater than or equal to x" list.

Time complexity: O(n) where n is the number of nodes in the linked list.
Space complexity: O(1) as we are not using any extra space apart from a few pointers.

Solution Code

public ListNode partition(ListNode head, int x) {
    ListNode lessThanX = new ListNode(0);
    ListNode greaterThanOrEqualX = new ListNode(0);
    ListNode lessThanXPtr = lessThanX;
    ListNode greaterThanOrEqualXPtr = greaterThanOrEqualX;
    
    while (head != null) {
        if (head.val < x) {
            lessThanXPtr.next = head;
            lessThanXPtr = lessThanXPtr.next;
        } else {
            greaterThanOrEqualXPtr.next = head;
            greaterThanOrEqualXPtr = greaterThanOrEqualXPtr.next;
        }
        head = head.next;
    }
    
    greaterThanOrEqualXPtr.next = null;
    lessThanXPtr.next = greaterThanOrEqualX.next;
    
    return lessThanX.next;
}

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