这题有几个关键的点
- 建立哨兵（sentinel）节点便于返回链表头
- 快慢指针：当快指针指向倒数第1个节点，慢指针必然指向倒数第1 + N个节点
- 节点是被new出来的，C++中需要手动释放（delete）这个节点的内存
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
// if n = 2
ListNode* removeNthFromEnd(ListNode* head, int n) {
// sentinel -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 ->
// slow
// fast
ListNode* sentinel = new ListNode(0);
sentinel->next = head;
ListNode* slow = sentinel;
ListNode* fast = sentinel;
// sentinel -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 ->
// slow fast
//
// Find Nth from end <-> fast is N steps faster than slow
for (int i = 1; i <= n; i++) {
fast = fast->next;
}
// sentinel -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 ->
// slow fast
while (fast->next != nullptr) {
slow = slow->next;
fast = fast->next;
}
// sentinel -> 1 -> 2 -> 3 -> 4 -> 6 ->
// slow fast
ListNode* temp = slow->next;
slow->next = slow->next->next;
delete temp;
return sentinel->next;
}
};
