Remove Nth Node From End of List

题目

Remove Nth Node From End of ListJan 28 '124532 / 11985

Given a linked list, remove the n^th 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

思路

快慢指针，时间复杂度为O(n) ，只扫描一遍。

/** * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(!head)            return head;        ListNode *cur = head;        for(int i=1;i<=n;i++)            cur = cur->next;        ListNode *pre = head;        while(cur && cur->next)        {            pre = pre->next;            cur = cur->next;        }        if(cur==NULL)        {            return pre->next;        }        pre->next = pre->next->next;        return head;            }};

 

最新 java

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode removeNthFromEnd(ListNode head, int n) {        if(head == null || n < 1){            return head;        }        ListNode dump = new ListNode(0);        dump.next = head;        ListNode slow = dump;        ListNode fast = head;        int i = 1;        while(fast != null && i++ < n){            fast = fast.next;        }        while(fast.next != null){            slow = slow.next;            fast = fast.next;        }        slow.next = slow.next.next;        return dump.next;    }}