问题
Given the head of a linked list, remove the nth node from the end of the list and return its head.
双指针,同时记录前n个节点和当前节点。
当前指针到链表尾部时,删除前面的指针,注意处理edge cases。
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class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { ListNode preNode = null; ListNode removedNode = head; ListNode fastNode = head; for ( int i = 0; i < n; i++ ){ fastNode = fastNode.next; } while ( fastNode != null ){ fastNode = fastNode.next; preNode = removedNode; removedNode = removedNode.next; } if ( removedNode == head ){ head = head.next; } else if ( removedNode.next == null){ preNode.next = null; } else{ preNode.next = removedNode.next; } return head; } }
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