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[LeetCode] 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.
思路一：快慢指针。时间复杂度O(n),空间复杂度O(1)
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution {10 public:11 ListNode *removeNthFromEnd(ListNode *head, int n) {12 if (head == NULL || n <= 0) return head;13 ListNode *pfast = head;14 for (int i = 1; i < n && pfast->next != NULL; ++i) {15 pfast = pfast->next;16 }17 ListNode **pslow = &head;18 while (pfast->next != NULL) {19 pfast = pfast->next;20 pslow = &((*pslow)->next);21 }22 23 ListNode *q = *pslow;24 *pslow = q->next;25 delete q;26 27 return head;28 }29 };
[LeetCode] Remove Nth Node From End of List
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