《leetCode》:Remove Nth Node From End of List
2017-08-07 20:07
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题目:
Given a linked list, remove the nth node from the end of list and return its head.
For example,
Note:
Given n will always be valid.
Try to do this in one pass.
题目大意:删除链表的倒数第n个结点。要注意的是:n值一直是有效的。好好品味这句话。
思路:
可以使用指针来完成。将一个指针快速移动——> n + 1位置向前,在两个指针之间保持n个间隔,然后以相同的速度移动。最后,当快速指针到达末端时,慢指针将会是n + 1的位置——正好是它能够跳过下一个节点的正确位置。
因为问题是,n是有效的,没有太多的检查必须到位。否则,这是必要的。
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode start = new ListNode(0);
ListNode slow = start, fast = start;
slow.next = head;
//Move fast in front so that the gap between slow and fast becomes n
for(int i=1; i<=n+1; i++) {
fast = fast.next;
}
//Move fast to the end, maintaining the gap
while(fast != null) {
slow = slow.next;
fast = fast.next;
}
//Skip the desired node
slow.next = slow.next.next;
return start.next;
}
以上
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.
题目大意:删除链表的倒数第n个结点。要注意的是:n值一直是有效的。好好品味这句话。
思路:
可以使用指针来完成。将一个指针快速移动——> n + 1位置向前,在两个指针之间保持n个间隔,然后以相同的速度移动。最后,当快速指针到达末端时,慢指针将会是n + 1的位置——正好是它能够跳过下一个节点的正确位置。
因为问题是,n是有效的,没有太多的检查必须到位。否则,这是必要的。
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode start = new ListNode(0);
ListNode slow = start, fast = start;
slow.next = head;
//Move fast in front so that the gap between slow and fast becomes n
for(int i=1; i<=n+1; i++) {
fast = fast.next;
}
//Move fast to the end, maintaining the gap
while(fast != null) {
slow = slow.next;
fast = fast.next;
}
//Skip the desired node
slow.next = slow.next.next;
return start.next;
}
以上
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