[LeetCode] 19. Remove Nth Node From End of List 移除链表倒数第N个节点

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.

移除给定链表的倒数第n个节点，n总是有效的，要求一次遍历完成one pass。

解法1: two pass，先求出链表的总长度，然后在删除node

解法2：双指针，经典题。主要是在一个遍历中找到第n个倒数元素，一个指针先走n步，然后两个指针同步走，直到第一个走到终点，第二个指针指向的就是需要删除的节点。Corner case: head == null; 头节点的处理，比如，1->2->NULL, n =2; 这时，要删除的就是头节点。

Java: Naive Two Pass

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public ListNode removeNthFromEnd(ListNode head, int n) {     if(head == null)         return null;        //get length of list     ListNode p = head;     int len = 0;     while(p != null){         len++;         p = p.next;     }        //if remove first node     int fromStart = len-n+1;     if(fromStart==1)         return head.next;        //remove non-first node        p = head;     int i=0;     while(p!=null){         i++;         if(i==fromStart-1){             p.next = p.next.next;         }         p=p.next;     }        return head; }

Java:

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public ListNode removeNthFromEnd(ListNode head, int n) {     if(head == null)         return null;        ListNode fast = head;     ListNode slow = head;        for(int i=0; i<n; i++){         fast = fast.next;     }        //if remove the first node     if(fast == null){         head = head.next;         return head;     }        while(fast.next != null){         fast = fast.next;         slow = slow.next;     }        slow.next = slow.next.next;        return head; }

Python:

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class ListNode:     def __init__(self, x):         self.val = x         self.next = None       def __repr__(self):         if self is None:             return "Nil"         else:             return "{} -> {}".format(self.val, repr(self.next))           class Solution:     # @return a ListNode     def removeNthFromEnd(self, head, n):         dummy = ListNode(-1)         dummy.next = head         slow, fast = dummy, dummy                   for i in xrange(n):             fast = fast.next                       while fast.next:             slow, fast = slow.next, fast.next                       slow.next = slow.next.next                   return dummy.next

C++:

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class Solution { public:     ListNode* removeNthFromEnd(ListNode* head, int n) {         if (!head->next) return NULL;         ListNode *pre = head, *cur = head;         for (int i = 0; i < n; ++i) cur = cur->next;         if (!cur) return head->next;         while (cur->next) {             cur = cur->next;             pre = pre->next;         }         pre->next = pre->next->next;         return head;     } };

　　

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