init impl

Author: freemanzhang

src/binarySearchTree/DeleteNodeInABST.java

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+package binarySearchTree;
+
+import utility.TreeNode;
+
+/*
+Given a root node reference of a BST and a key, delete the node with the given key in the BST. 
+Return the root node reference (possibly updated) of the BST.
+
+Basically, the deletion can be divided into two stages:
+
+Search for a node to remove.
+If the node is found, delete the node.
+Note: Time complexity should be O(height of tree).
+
+Example:
+
+root = [5,3,6,2,4,null,7]
+key = 3
+
+    5
+   / \
+  3   6
+ / \   \
+2   4   7
+
+Given key to delete is 3. So we find the node with value 3 and delete it.
+
+One valid answer is [5,4,6,2,null,null,7], shown in the following BST.
+
+    5
+   / \
+  4   6
+ /     \
+2       7
+
+Another valid answer is [5,2,6,null,4,null,7].
+
+    5
+   / \
+  2   6
+   \   \
+    4   7
+ * */
+
+public class DeleteNodeInABST
+{
+    public TreeNode deleteNode( TreeNode root, int key )
+    {
+    	if ( root == null )
+    	{
+    		return root;
+    	}
+    	if ( root.val > key )
+    	{
+    		root.left = deleteNode( root.left, key );
+    	}
+    	else if ( root.val < key )
+    	{
+    		root.right = deleteNode( root.right, key );
+    	}
+    	else
+    	{
+    		if ( root.left == null )
+    		{
+    			return root.right;
+    		}
+    		else if ( root.right == null )
+    		{
+    			return root.left;
+    		}
+    		else // two children
+    		{
+    			root.val = getMin( root.right );
+    			root.right = deleteNode( root.right, root.val );
+    		}
+    	}
+    	return root;
+    }
+    
+    private int getMin( TreeNode root )
+    {
+    	int min = root.val;
+    	while ( root.left != null )
+    	{
+    		root = root.left;
+    		min = root.val;
+    	}
+    	return min;
+    }
+}

src/hashtable/SortCharactersByFrequency.java

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+package hashtable;
+
+import java.util.HashMap;
+import java.util.Map;
+import java.util.PriorityQueue;
+
+/*
+Given a string, sort it in decreasing order based on the frequency of characters.
+
+Example 1:
+
+Input:
+"tree"
+
+Output:
+"eert"
+
+Explanation:
+'e' appears twice while 'r' and 't' both appear once.
+So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.
+Example 2:
+
+Input:
+"cccaaa"
+
+Output:
+"cccaaa"
+
+Explanation:
+Both 'c' and 'a' appear three times, so "aaaccc" is also a valid answer.
+Note that "cacaca" is incorrect, as the same characters must be together.
+Example 3:
+
+Input:
+"Aabb"
+
+Output:
+"bbAa"
+
+Explanation:
+"bbaA" is also a valid answer, but "Aabb" is incorrect.
+Note that 'A' and 'a' are treated as two different characters.
+ * */
+
+public class SortCharactersByFrequency
+{
+    public String frequencySort( String s )
+    {
+    	if ( s == null || s.length() == 0 )
+    	{
+    		return "";
+    	}
+    	
+    	Map<Character, Integer> histogram = new HashMap<>();
+    	for ( Character ch : s.toCharArray() )
+    	{
+    		histogram.put( ch, histogram.getOrDefault( ch, 0 ) + 1 );
+    	}
+    	
+    	PriorityQueue<Map.Entry<Character,Integer>> maxQueue = new PriorityQueue<>( ( o1, o2 ) -> o2.getValue() - o1.getValue() );
+    	for ( Map.Entry<Character, Integer> entry : histogram.entrySet() )
+    	{
+    		maxQueue.offer( entry );
+    	}
+    	
+    	StringBuilder result = new StringBuilder();
+    	while ( !maxQueue.isEmpty() )
+    	{
+    		Map.Entry<Character, Integer> entry = maxQueue.poll();
+    		for ( int i = 0; i < entry.getValue(); i++ )
+    		{
+    			result.append( entry.getKey() );
+    		}
+    	}
+    	return result.toString();
+    }    
+}

src/string/RightRotation.java

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+package string;
+
+/*
+判断string1和string2是否互为right rotation。比如：abcd和cdab。CC150原题， 
+一行代码即可解决。 
+ * */
+
+public class RightRotation
+{
+	public static int rightRotate( String word1, String word2 )
+	{
+		if ( word1 == null || word2 == null || word1.length() == 0 || word2.length() == 0
+				|| word1.length() != word2.length() )
+		{
+			return -1;
+		}
+		String str = word1 + word1;
+		return str.indexOf( word2 ) != -1 ? 1 : -1;
+	}
+}