Leetcode - 19. Remove Nth Node from end of List

问题简介

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.

解法一

/**
* 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) {
int length = 0;
ListNode* hd = head;
ListNode* newHead = head;

// 考虑1个元素 eg:[1] 1
if(head->next==NULL)return NULL;

while(head){
head = head->next;
length++;
}
// 考虑除去第一个元素
if(n==length)
{
hd=hd->next;
return hd;
}
// 考虑去除中间元素
n = length-n;
for(int i=0;i<n-1;i++){
hd = hd->next;
}

hd->next = hd->next->next;
return newHead;
}
};

解法二

引入newHead(-1)是为了保证在删除首元素时避免循环的first->next不存在。eg: N为原链表长度时，first遍历后first == NULL

由于开头加了一个元素，first指针在遍历N步后

引入second指针随着first的遍历完全，指向被删除元素的前一个位置。

最后，second->next = second->next->next;

class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* newHead = new ListNode(-1);
newHead->next = head;

ListNode*first = newHead;
ListNode*second = newHead;

for(int i=0;i<n;i++){
first = first->next;
}

while(first->next){
first = first->next;
second = second->next;
}

ListNode* tmp = second->next;
second->next = second->next->next;
delete tmp;

return newHead->next;
}