LeetCode-19. Remove Nth Node From End of List

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.

删除链表的倒数第N个元素，只能扫描一遍

package solutions._19;

import utils.ListNode;

/**
* 19. Remove Nth Node From End of List
*/

class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode slow = head;
ListNode fast = head;
for (int i = 0; i < n; i++) {
fast = fast.next;
}

// 删除第一个元素
if (fast == null) {
// 存在多个元素
if (slow.next != null) {
slow.val = slow.next.val;
slow.next = slow.next.next;
return head;
} else {
// 只存在一个元素
return null;
}

}

while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}

// 因为找到的 slow 代表要删除的元素的前一个元素
// 所以 slow.next 一定不为空
if (slow.next.next == null) {
slow.next = null;
} else {
slow.next = slow.next.next;
}
return head;
}

public static void main(String[] args) {
ListNode node1 = new ListNode(1);
ListNode node2 = new ListNode(2);
ListNode node3 = new ListNode(3);
ListNode node4 = new ListNode(4);
ListNode node5 = new ListNode(5);
ListNode node6 = new ListNode(6);

node1.next = node2;
node2.next = null;
node3.next = node4;
node4.next = node5;
node5.next = node6;
node6.next = null;

Solution solution = new Solution();
ListNode p = solution.removeNthFromEnd(node1, 1);
while (p != null) {
System.out.print(p.val + " ");
p = p.next;
}
}
}

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