Programming
Leetcode-19. Remove Nth Node From End of List
Given the head of a linked list, remove the nth node from the end of the list and return its head.
Example 1:
Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]
Example 2:
Input: head = [1], n = 1
Output: []
Example 3:
Input: head = [1,2], n = 1
Output: [1]
<解題>
- 使用 fast 指針向前移動 n + 1 步，以創造出 n 個節點的間隔，這樣在 fast 達到鏈表尾部時，slow 將指向倒數第 n 個節點的前一個節點
- 最後指向倒數n的下一位
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0);
ListNode slow = dummy, fast = dummy;
dummy.next = head;
//先讓fast走n步
for(int i=0; i<=n; i++) {
fast = fast.next;
}
//slow會走到倒數n的位置
while(fast != null) {
slow = slow.next;
fast = fast.next;
}
//跳過倒數n的下一位
dummy.next = slow.next.next;
return dummy.next;
}
}