LeetCode 19: Remove Nth Node From End of List
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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; }};
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