(算法分析Week16)Remove Nth Node From End of List[Medium]

309. 19. Remove Nth Node From End of List[Medium]

题目来源

Description

Given a linked list, remove the nth node from the end of list and return its head.

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.

Solution

挺简单的一道题。删除链表哦中倒数第n个数，要求只能遍历一次，所以不能先遍历一次看共有多少个元素再重新遍历到size-n。

用两个指针，一个指针先走n步，此时如果把尾看成头，则相当于走到了倒数第n+1个数的位置。另一个指针从这时候开始，两个指针同时移动，当第一个指针移动到尾节点的时候，第二个指针的next就是要被删除的节点。

需要注意的是，如果第一个指针走完n步已经到尾节点，说明删除的就是head，返回head->next即可。

Complexity analysis

O(n)

Code

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
if (head -> next == NULL || head == NULL)
return NULL;
ListNode* p = head;
for (int i = 0; i < n; i++) {
p = p -> next;
}
if (p == NULL)
return head -> next;
else {
ListNode* q = head;
while(p->next != NULL) {
q = q -> next;
p = p -> next;
}
ListNode* temp = q -> next;
q -> next = temp -> next;
delete temp;
}
return head;
}
};

Result