19. Remove Nth Node From End of List

Given a linked list, remove the n^th node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

官方答案如下：

Algorithm

The above algorithm could be optimized to one pass. Instead of one pointer, we could use two pointers. The first pointer advances the list by n+1n+1n+1 steps from the beginning, while the second pointer starts from the beginning of the list. Now, both pointers are exactly separated by nnn nodes apart. We maintain this constant gap by advancing both pointers together until the first pointer arrives past the last node. The second pointer will be pointing at the nnnth node counting from the last.We relink the next pointer of the node referenced by the second pointer to point to the node's next next node.

19. Remove Nth Node From End of List

Figure 2. Remove the nth element from end of a list.

Java

public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode first = dummy;
    ListNode second = dummy;
    // Advances first pointer so that the gap between first and second is n nodes apart
    for (int i = 1; i <= n + 1; i++) {
        first = first.next;
    }
    // Move first to the end, maintaining the gap
    while (first != null) {
        first = first.next;
        second = second.next;
    }
    second.next = second.next.next;
    return dummy.next;
}

Complexity Analysis

Time complexity : O(L)O(L)O(L). The algorithm makes one traversal of the list of LLL nodes. Therefore time complexity is O(L)O(L)O(L).
Space complexity : O(1)O(1)O(1).We only used constant extra space.

Analysis written by: @elmirap.