LeetCode 19: Remove Nth Node From End of List

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.

解题思路

使用两个指针（Two Pointers） p1 和 p2 指向表头结点，让 p2 先走 n 步，然后 p1 和 p2 同时向前移动，直到 p2 为空，则 p1 指向要删除的结点。在此过程中保存 p1 的前驱结点信息。代码如下：

/**
* 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) {
// 使用两个指针p1，p2；p1Pre指向p1的前驱
ListNode *p1Pre = NULL, *p1 = head, *p2 = head;

// 让p2先走n步
for (int i = 0; i < n; ++i) {
p2 = p2->next;
}

// p1和p2同时向前移动，直到p2为空，则p1指向要删除的结点
while (p2) {
p1Pre = p1;
p1 = p1->next;
p2 = p2->next;
}

// 删除p1，注意p1可能指向头结点
if (p1Pre) {
p1Pre->next = p1->next;
}
else {
head = p1->next;
}

delete p1;
return head;
}
};