【試題描述】定義一個函數，輸入一個鏈表，刪除無序鏈表中重復的節點

【參考代碼】

方法一：

Without a buffer, we can iterate with two pointers: “current” does a normal iteration, while?
“runner” iterates through all prior nodes to check for dups Runner will only see one dup?
per node, because if there were multiple duplicates they would have been removed already

1 public static void deleteDups(LinkList head) 2 { 3 if (head == null) 4 return; 5 Link previous = head.first; 6 Link current = previous.next; 7 while (current != null) 8 { 9 Link runner = head.first; 10 while (runner != current) 11 { 12 if (runner.id == current.id) 13 { 14 Link tmp = current.next; 15 previous.next = tmp; 16 current = tmp; 17 break; 18 } 19 runner = runner.next; 20 } 21 22 if (runner == current) 23 { 24 previous = current; 25 current = current.next; 26 } 27 } 28 29 System.out.println("-----------"); 30 head.displayList(); 31 }

方法二：

If we can use a buffer, we can keep track of elements in a hashtable and remove any dups:

public static void deleteDups2(LinkList head){if (head == null)return;Hashtable table = new Hashtable();Link previous = null;Link current = head.first;while (current != null){if (table.containsKey(current.id))previous.next = current.next;else{table.put(current.id, true);previous = current;}current = current.next;}System.out.println("-----------");head.displayList();}

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總結

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