Data Structures Algorithms 简明教程

1. Start
2. Create an Array of a desired datatype and size.
3. Initialize a variable ‘i’ as 0.
4. Enter the element at ith index of the array.
5. Increment i by 1.
6. Repeat Steps 4 & 5 until the end of the array.
7. Stop

1. Start
2. Set J = K
3. Repeat steps 4 and 5 while J < N
4. Set LA[J] = LA[J + 1]
5. Set J = J+1
6. Set N = N-1
7. Stop

1. Start
2. Set J = 0
3. Repeat steps 4 and 5 while J < N
4. IF LA[J] is equal ITEM THEN GOTO STEP 6
5. Set J = J +1
6. PRINT J, ITEM
7. Stop

1 Start
2. Initialize an Array of certain size and datatype.
3. Initialize another variable ‘i’ with 0.
4. Print the ith value in the array and increment i.
5. Repeat Step 4 until the end of the array is reached.
6. End

1. Start
2. Set LA[K-1] = ITEM
3. Stop

1. Start
2. Print all the elements in the Array
3. Stop

1. START
2. Create a node to store the data
3. Check if the list is empty
4. If the list is empty, add the data to the node and assign the head pointer to it.
5 If the list is not empty, add the data to a node and link to the current head. Assign the head to the newly added node.
6. END

1. START
2. Create a new node and assign the data
3. Find the last node
4. Point the last node to new node
5. END

1. START
2. Create a new node and assign data to it
3. Iterate until the node at position is found
4. Point first to new first node
5. END

1. START
2. Assign the head pointer to the next node in the list
3. END

1. START
2. Iterate until you find the second last element in the list.
3. Assign NULL to the second last element in the list.
4. END

1. START
2. Iterate until find the current node at position in the list
3. Assign the adjacent node of current node in the list to its previous node.
4. END

1 START
2. We use three pointers to perform the reversing: prev, next, head.
3. Point the current node to head and assign its next value to the prev node.
4. Iteratively repeat the step 3 for all the nodes in the list.
5. Assign head to the prev node.

1 START
2 If the list is not empty, iteratively check if the list contains the key
3 If the key element is not present in the list, unsuccessful search
4 END

1. START
2. While the list is not empty and did not reach the end of the list, print the data in each node
3. END

1. START
2. Create a new node with three variables: prev, data, next.
3. Store the new data in the data variable
4. If the list is empty, make the new node as head.
5. Otherwise, link the address of the existing first node to the next variable of the new node, and assign null to the prev variable.
6. Point the head to the new node.
7. END

1. START
2. Check the status of the doubly linked list
3. If the list is empty, deletion is not possible
4. If the list is not empty, the head pointer is shifted to the next node.
5. END

1. START
2. If the list is empty, add the node to the list and point the head to it.
3. If the list is not empty, find the last node of the list.
4. Create a link between the last node in the list and the new node.
5. The new node will point to NULL as it is the new last node.
6. END

1. START
2. Check if the list is empty
3. If the list is empty, add the node and point the head to this node
4. If the list is not empty, link the existing head as the next node to the new node.
5. Make the new node as the new head.
6. END

1. START
2. If the list is empty, then the program is returned.
3. If the list is not empty, we traverse the list using a current pointer that is set to the head pointer and create another pointer previous that points to the last node.
4. Suppose the list has only one node, the node is deleted by setting the head pointer to NULL.
5. If the list has more than one node and the first node is to be deleted, the head is set to the next node and the previous is linked to the new head.
6. If the node to be deleted is the last node, link the preceding node of the last node to head node.
7. If the node is neither first nor last, remove the node by linking its preceding node to its succeeding node.
8. END

1. START
2. Walk through all the nodes of the list and print them
3. END

1 − Checks if the stack is full.
2 − If the stack is full, produces an error and exit.
3 − If the stack is not full, increments top to point next empty space.
4 − Adds data element to the stack location, where top is pointing.
5 − Returns success.

1 − Checks if the stack is empty.
2 − If the stack is empty, produces an error and exit.
3 − If the stack is not empty, accesses the data element at which top is pointing.
4 − Decreases the value of top by 1.
5 − Returns success.

1. START
2. return the element at the top of the stack
3. END

1. START
2. If the size of the stack is equal to the top position of the stack, the stack is full. Return 1.
3. Otherwise, return 0.
4. END

1. START
2. If the top value is -1, the stack is empty. Return 1.
3. Otherwise, return 0.
4. END

1 − START
2 – Check if the queue is full.
3 − If the queue is full, produce overflow error and exit.
4 − If the queue is not full, increment rear pointer to point the next empty space.
5 − Add data element to the queue location, where the rear is pointing.
6 − return success.
7 – END

1 – START
2 − Check if the queue is empty.
3 − If the queue is empty, produce underflow error and exit.
4 − If the queue is not empty, access the data where front is pointing.
5 − Increment front pointer to point to the next available data element.
6 − Return success.
7 – END

1 – START
2 – Return the element at the front of the queue
3 – END

1 – START
2 – If the count of queue elements equals the queue size, return true
3 – Otherwise, return false
4 – END

1 – START
2 – If the count of queue elements equals zero, return true
3 – Otherwise, return false
4 – END

1. START
2. Check whether the tree is empty or not
3. If the tree is empty, search is not possible
4. Otherwise, first search the root of the tree.
5. If the key does not match with the value in the root, search its subtrees.
6. If the value of the key is less than the root value, search the left subtree
7. If the value of the key is greater than the root value, search the right subtree.
8. If the key is not found in the tree, return unsuccessful search.
9. END

1 – START
2 – If the tree is empty, insert the first element as the root node of the tree. The following elements are added as the leaf nodes.
3 – If an element is less than the root value, it is added into the left subtree as a leaf node.
4 – If an element is greater than the root value, it is added into the right subtree as a leaf node.
5 – The final leaf nodes of the tree point to NULL values as their child nodes.
6 – END

1. START
2. Traverse the left subtree, recursively
3. Then, traverse the root node
4. Traverse the right subtree, recursively.
5. END

1. START
2. Traverse the root node first.
3. Then traverse the left subtree, recursively
4. Later, traverse the right subtree, recursively.
5. END

1. START
2. Traverse the left subtree, recursively
3. Traverse the right subtree, recursively.
4. Then, traverse the root node
5. END

START
   if node == null then:
      return new node
   if key < node.key then:
      node.left = insert (node.left, key)
   else if (key > node.key) then:
      node.right = insert (node.right, key)
   else
      return node
   node.height = 1 + max (height (node.left), height (node.right))
   balance = getBalance (node)
   if balance > 1 and key < node.left.key then:
      rightRotate
   if balance < -1 and key > node.right.key then:
      leftRotate
   if balance > 1 and key > node.left.key then:
      node.left = leftRotate (node.left)
      rightRotate
   if balance < -1 and key < node.right.key then:
      node.right = rightRotate (node.right)
      leftRotate (node)
   return node
END

START
   if root == null: return root
   if key < root.key:
      root.left = delete Node
   else if key > root.key:
      root.right = delete Node
   else:
      if root.left == null or root.right == null then:
         Node temp = null
         if (temp == root.left)
            temp = root.right
         else
            temp = root.left
         if temp == null then:
            temp = root
            root = null
         else
            root = temp
         else:
            temp = minimum valued node
            root.key = temp.key
            root.right = delete Node
      if (root == null) then:
         return root
         root.height = max (height (root.left), height (root.right)) + 1
         balance = getBalance
      if balance > 1 and getBalance (root.left) >= 0:
         rightRotate
      if balance > 1 and getBalance (root.left) < 0:
         root.left = leftRotate (root.left);
         rightRotate
      if balance < -1 and getBalance (root.right) <= 0:
         leftRotate
      if balance < -1 and getBalance (root.right) > 0:
         root.right = rightRotate (root.right);
         leftRotate
      return root
END

// Insert the value
void insertion(int item) {
   int flag, i;
   struct btreeNode *child;
   flag = setNodeValue(item, &i, root, &child);
   if (flag)
      root = createNode(i, child);
}

// Deletion operation
void deletion(int key){
   int index = searchkey(key);
   if (index < n && keys[index] == key) {
      if (leaf)
         deletion_at_leaf(index);
      else
         deletion_at_nonleaf(index);
   } else {
      if (leaf) {
         cout << "key " << key << " does not exist in the tree\n";
         return;
      }
      bool flag = ((index == n) ? true : false);
      if (C[index]->n < t)
         fill(index);
      if (flag && index > n)
         C[index - 1]->deletion(key);
      else
         C[index]->deletion(key);
   }
   return;
}

// Deletion at the leaf nodes
void deletion_at_leaf(int index){
   for (int i = index + 1; i < n; ++i)
      keys[i - 1] = keys[i];
   n--;
   return;
}

// Deletion at the non leaf node
void deletion_at_nonleaf(int index){
   int key = keys[index];
   if (C[index]->n >= t) {
      int pred = get_Predecessor(index);
      keys[index] = pred;
      C[index]->deletion(pred);
   } else if (C[index + 1]->n >= t) {
      int successor = copysuccessoressor(index);
      keys[index] = successor;
      C[index + 1]->deletion(successor);
   } else {
      merge(index);
      C[index]->deletion(key);
   }
   return;
}