I was trying to solve the Partition List problem on LeetCode. The problem asks to sort an linked list given a target list node, so that all the list nodes having smaller value than the target will come before the target, while their original relative order remains unchanged.

I was able to come up a straightforward algorithm and to pass the online judge, basically creating two pointers and using them each to link either nodes < target or >= target while traversing the list.

public ListNode partition(ListNode head, int x) {
    ListNode ptr = head;
    ListNode small = new ListNode(0);
    ListNode big = new ListNode(0);
    ListNode dummy_1 = big;
    ListNode dummy_2 = small;
    int i = 1;
    while (ptr != null) {
        if (ptr.val < x) {
          small.next = ptr;
          small = small.next;
        } else {
          big.next = ptr;
          big = big.next;
        }
        ListNode help = ptr.next;
        ptr.next = null;
        ptr = help;
    }
    small.next = dummy_1.next;
    return dummy_2.next;
}

the following codes breaks the the link between ptr and ptr.next, and move the
ptr to original ptr.next.

ListNode help = ptr.next;
ptr.next = null;
ptr = help;

What I haven't quite figured out yet is that why this step is necessary, as we can move ptr to its next and directly update the reference later using small.next = ptr and big.next = ptr in the while-loop;

However when I simply used ptr = ptr.next instead of the three lines of code above, the online judge responded with error Memory Limit Exceeded.

I would really appreciate if anyone can explain this for me. What may cause the Memory Limit error as any cycled-list has seemed to be already avoided?

As commented, just setting big.next = null is working (I ran this using netbeans / java).

static ListNode partition(ListNode head, int x) {
    ListNode ptr = head;
    ListNode small = new ListNode(0);
    ListNode big = new ListNode(0);
    ListNode dummy_1 = big;
    ListNode dummy_2 = small;
    while (ptr != null) {
        if (ptr.val < x) {
          small.next = ptr;
          small = small.next;
        } else {
          big.next = ptr;
          big = big.next;
        }
        ptr = ptr.next;
    }
    small.next = dummy_1.next;
    big.next = null;
    return dummy_2.next;
}

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