LeetCode 16 Remove Nth Node From End of List

方法一： 双指针容易点。

public class Solution {
  public ListNode removeNthFromEnd(ListNode head, int n) {
    // Write your solution here
    if (head == null) {
      return head; 
    }
    
    if (n > getNum(head)) {
     return head; 
    }
    
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode fast = dummy, slow = dummy;
    for (int i = 0; i <= n; i++) {
      fast = fast.next; 
    }
    
    while (fast != null) {
      fast = fast.next;
      slow = slow.next;
    }
    slow.next = slow.next.next;
    return dummy.next;
  }
  
  private int getNum(ListNode head) {
    int count = 0;
    while (head != null) {
      count++;
      head = head.next; 
    }
    return count;
  }
}

方法二： 数个数

public ListNode removeNthNode(ListNode head, int n) {
// Should using dummy node here because n may remove the head node.So we must keep the head.
if (head == null || n == 0) {
    return head;
}

ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode cur = dummy;

ListNode fast = head;
ListNode slow = head;

// Firstly, count the length of the linked list
int len = 1;
while(fast.next != null && fast.next.next != null) {
    fast = fast.next.next;
    slow = slow.next;
    len++;
}
//          1  2  3  4  5  6  7  8 null   , for n = 5
//  dummy             n
if (fast.next == null) {
    len = 2 * len - 1;
} else if (fast.next.next == null) {
    len = 2 * len;
}
    
// Calculate the difference of length: 8 - 5 = 3
int pos = len - n;

// If we use the dummy node, then the i must start from i = 0;
for (int i = 0; i < pos; i++) {
    cur = cur.next;
}
// cur points to 3, we need to delete 4
cur.next = cur.next.next;
return dummy.next;
}