LeetCode-Remove Nth Node From End of List
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题目:
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.
本题的解法是基于双指针的,保持两个指针的距离为n,比较快的指针到达末尾了,那么比较慢的指针所指向的就是要删除的点。当然距离最好为n+1,这样比较慢的指针指向的为前趋,比较容易操作删除。代码如下:
public ListNode removeNthFromEnd(ListNode head, int n) { if (n <= 0) { return head; } ListNode pre = head; ListNode cur = head; int i = 0; while (i <= n && cur != null) { cur = cur.next; i++; } if (i <= n) { return pre.next; } while (cur != null) { pre = pre.next; cur = cur.next; } pre.next = pre.next.next; return head; }
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