Beispiel #1
37
  public static void main(String[] args) {
    ListNode a = new ListNode('1');
    ListNode b = new ListNode('2');
    ListNode c = new ListNode('3');
    ListNode d = new ListNode('1');
    ListNode e = new ListNode(' ');
    ListNode f = new ListNode('3');

    a.next = b;
    b.next = c;
    c.next = d;
    d.next = e;
    e.next = f;

    System.out.println(a);
    removeDupes(a);
    System.out.println("After removing duplicate ListNodes: ");
    System.out.println(a);
  }
Beispiel #2
1
 public ListNode partition(ListNode head, int x) {
   ListNode headOfHead = new ListNode(-1);
   headOfHead.next = head;
   ListNode p = headOfHead;
   while (p.next != null) {
     p = p.next;
     // CUT begin
     debug(p.val);
     // CUT end
   }
   ListNode tail = new ListNode(-1);
   p.next = tail;
   p = headOfHead;
   ListNode q = tail;
   while (p.next != tail) {
     if (p.next.val < x) {
       p = p.next;
     } else {
       q.next = p.next;
       q = q.next;
       p.next = p.next.next;
       q.next = null;
     }
   }
   p.next = tail.next;
   return headOfHead.next;
 }
Beispiel #3
0
 public ListNode deleteDuplicates(ListNode a) {
   ListNode temp = a;
   LinkedHashMap<Integer, Integer> map = new LinkedHashMap<Integer, Integer>();
   int i = 0;
   while (temp != null) {
     if (map.get(temp.val) == null) {
       map.put(temp.val, 1);
     } else {
       map.put(temp.val, map.get(temp.val) + 1);
     }
     temp = temp.next;
   }
   System.out.println(map);
   Iterator<Integer> it = map.keySet().iterator();
   ListNode root = null;
   ListNode root1 = null;
   while (it.hasNext()) {
     int x = it.next();
     if (map.get(x) == 1) {
       if (root == null) {
         root = new ListNode(x);
         root1 = root;
       } else {
         ListNode node = new ListNode(x);
         root1.next = node;
         root1 = node;
       }
     }
   }
   return root;
 }
Beispiel #4
0
  /*
  Solution 2
  	Did not use additional space, each time encounter the first unique value node,
  	traversal the rest of the list to delete the same value's nodes.
  Assumption:
  	None.
  Time complexity: O(n^2)
  Space Complexity: O(1)
  */
  private static ListNode RemoveDups2(ListNode head) {
    if (head == null) return head;

    ListNode pre = head;
    while (pre != null) {
      ListNode pre2 = pre;
      ListNode cur = pre.next;
      int target = pre.val;

      while (cur != null) { // delete the dups in the rest of the list
        if (cur.val == target) {
          pre2.next = cur.next;
          cur = cur.next;
        } else {
          pre2 = cur;
          cur = cur.next;
        }
      }

      // move to next different value's node
      pre = pre.next;
    }

    return head;
  }
Beispiel #5
0
  /* Helpers */
  private static InputArgs input() throws IOException {
    System.out.println("2.2 Return Kth to Last");

    // read K
    System.out.print("Please input K: ");
    Scanner sc = new Scanner(System.in);
    int k = sc.nextInt();

    // read number of nodes
    System.out.println();
    System.out.print("Please input the number of nodes in the linked list: ");
    int n = sc.nextInt();

    // read linked list
    System.out.println();
    System.out.print("Please input value of nodes (seperate by space, e.g., 1 2 3): ");

    ListNode dummy = new ListNode(0);
    ListNode pre = dummy;
    for (int i = 0; i < n; i++) {
      ListNode cur = new ListNode(sc.nextInt());
      pre.next = cur;
      pre = cur;
    }

    return new InputArgs(k, dummy.next);
  }
  public int search(Object p) {
    int index = 0;
    boolean flag = false;
    temp = back;
    ListNode temp2 = new ListNode(p, null);

    if (back == null) {
      return -1;
    }
    // else if case

    else {
      while (temp.getNext() != null && flag == false) {
        if (temp.getValue() == temp2.getValue()) {
          flag = true;
        } else {
          temp = temp.getNext();
          index++;
        }
      }

      if (flag) {
        return index;
      } else {
        return -1;
      }
    }
  }
 public ListNode removeLast() {
   if (front == null) {
     return null;
   } else {
     ListNode temp2 = back;
     while (temp2.getNext() != front && temp2.getNext() != null) // second condition test
     {
       temp2 = temp2.getNext();
     }
     temp = front;
     front = temp2;
     size--;
     return temp;
   }
 }
Beispiel #8
0
 public void solve(InputReader in, OutputWriter out) {
   int n = in.nextInt();
   int x = in.nextInt();
   ListNode head = new ListNode(-1);
   ListNode p = head;
   for (int i = 0; i < n; i++) {
     p.next = new ListNode(in.nextInt());
     p = p.next;
     //            out.println(p.val);
   }
   ListNode res = new A1().partition(head.next, x);
   p = res;
   while (p != null) {
     out.println(p.val);
     p = p.next;
   }
 }
Beispiel #9
0
  public static void removeDupes(ListNode ListNode) {
    Set<Integer> set = new HashSet<Integer>();
    // add the first ListNode
    set.add(ListNode.val);

    ListNode prev = ListNode;
    ListNode curr = ListNode.next;
    while (curr != null) {
      // already exists, delete this duplicate one
      if (set.contains(curr.val)) {
        prev.next = curr.next;
      } else {
        set.add(curr.val);
        prev = curr;
      }
      curr = curr.next;
    }
  }
 public void insertLast(ListNode x) {
   if (front == null) {
     front = back = x;
     size++;
   } else {
     front.setNext(x);
     front = x;
     size++;
   }
 }
 public void insertFirst(ListNode x) {
   if (back == null) {
     front = back = temp = x;
     size++;
   } else {
     x.setNext(back);
     back = x;
     size++;
   }
 }
Beispiel #12
0
  /*
  Solution 1
  	Using hashtable to record nodes, delete the node if the same value already exists.
  Assumption:
  	None.
  Time complexity: O(n)
  Space Complexity: O(n)
  */
  private static ListNode RemoveDups1(ListNode head) {
    if (head == null) return head;

    Set<Integer> set = new HashSet<Integer>();
    ListNode pre = head;
    ListNode cur = head.next;
    set.add(head.val);
    while (cur != null) {
      if (set.contains(cur.val)) { // delete dup
        pre.next = cur.next;
        cur = cur.next;
      } else {
        set.add(cur.val);
        pre = cur;
        cur = cur.next;
      }
    }

    return head;
  }
  public ListNode removeFirst() {

    if (back == null) {
      return null;
    } else {
      temp = back;
      back = back.getNext();
      size--;
      return temp;
    }
  }
Beispiel #14
0
  // input
  private static ListNode input() {
    ListNode dummyhead = new ListNode(0);
    ListNode pre = dummyhead;

    System.out.println("/**** 2.1 Remove Dups ****/");
    System.out.println("Please input a linked list (using space to seperate nodes): ");

    try {
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      String s = br.readLine();
      String[] nodes = s.split(" ");
      for (int i = 0; i < nodes.length; i++) {
        ListNode cur = new ListNode(Integer.parseInt(nodes[i]));
        pre.next = cur;
        pre = cur;
      }
    } catch (IOException e) {
      System.out.println("input error, " + e.getMessage());
    }

    return dummyhead.next;
  }