Remove Nth Node From End of List
Source
Given a linked list, remove the nth node from the end of list and return its head. Note The minimum number of nodes in list is n. Example Given linked list: 1->2->3->4->5->null, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5->null. Challenge O(n) time
题解
简单题,使用快慢指针解决此题,需要注意最后删除的是否为头节点。让快指针先走n
步,直至快指针走到终点,找到需要删除节点之前的一个节点,改变node->next
域即可。
C++
/** * Definition of ListNode * class ListNode { * public: * int val; * ListNode *next; * ListNode(int val) { * this->val = val; * this->next = NULL; * } * } */ class Solution { public: /** * @param head: The first node of linked list. * @param n: An integer. * @return: The head of linked list. */ ListNode *removeNthFromEnd(ListNode *head, int n) { if (NULL == head || n < 1) { return NULL; } ListNode *dummy = new ListNode(0); dummy->next = head; ListNode *preDel = dummy; for (int i = 0; i != n; ++i) { if (NULL == head) { return NULL; } head = head->next; } while (head) { head = head->next; preDel = preDel->next; } preDel->next = preDel->next->next; return dummy->next; } };
Java
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { if (head == nul) return head; ListNode dummy = new ListNode(0); dummy.next = head; ListNode fast = head; ListNode slow = dummy; for (int i = 0; i < n; i++) { fast = fast.next; } while(fast != null) { fast = fast.next; slow = slow.next; } // gc friendly // ListNode toBeDeleted = slow.next; slow.next = slow.next.next; // toBeDeleted.next = null; // toBeDeleted = null; return dummy.next; } }
源码分析
引入dummy节点后画个图分析下就能确定head和preDel的转移关系了。 注意 while 循环中和快慢指针初始化的关系,否则容易在顺序上错一。
复杂度分析
极限情况下遍历两遍链表,时间复杂度为 O(n).