给定一个链表,删除链表的倒数第 n 个节点,并且返回链表的头结点。
示例:
给定一个链表: 1->2->3->4->5, 和 n = 2.
当删除了倒数第二个节点后,链表变为 1->2->3->5.
说明:
给定的 n 保证是有效的。
进阶:
你能尝试使用一趟扫描实现吗?
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
C++
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* dummyHead = new ListNode(0);
dummyHead->next = head;
ListNode* p = dummyHead;
ListNode* q = dummyHead;
for( int i = 0 ; i < n + 1 ; i ++ ){
q = q->next;
}
while(q){
p = p->next;
q = q->next;
}
ListNode* delNode = p->next;
p->next = delNode->next;
delete delNode;
ListNode* retNode = dummyHead->next;
delete dummyHead;
return retNode;
}
};
java1
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
int length = 0;
ListNode first = head;
while (first != null) {
length++;
first = first.next;
}
length -= n;
first = dummy;
while (length > 0) {
length--;
first = first.next;
}
first.next = first.next.next;
return dummy.next;
}
作者:LeetCode
链接:https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/solution/shan-chu-lian-biao-de-dao-shu-di-nge-jie-dian-by-l/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
java2
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode first = dummy;
ListNode second = dummy;
// Advances first pointer so that the gap between first and second is n nodes apart
for (int i = 1; i <= n + 1; i++) {
first = first.next;
}
// Move first to the end, maintaining the gap
while (first != null) {
first = first.next;
second = second.next;
}
second.next = second.next.next;
return dummy.next;
}
作者:LeetCode
链接:https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/solution/shan-chu-lian-biao-de-dao-shu-di-nge-jie-dian-by-l/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
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