Remove Nth Node From End of List
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Question Solution
Given a linked list, remove the nth 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.
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SOLUTION 1:
1.使用快慢指针，快指针先行移动N步。用慢指针指向要移除的Node的前一个Node.
2. 使用dummy node作为head的前缀节点，这样就算是删除head也能轻松handle啦！
主页君是不是很聪明呀？ :)
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
//
ListNode dummy = new ListNode();
dummy.next = head; ListNode slow = dummy;
ListNode fast = dummy; // move fast N more than slow.
while (n > ) {
fast = fast.next;
// Bug 1: FORGET THE N--;
n--;
} while (fast.next != null) {
fast = fast.next;
slow = slow.next;
} // Slow is the pre node of the node which we want to delete.
slow.next = slow.next.next; return dummy.next;
}
}
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