xiren-wang

## remove_duplicates_in_sorted_array_II.cpp
class Solution {
public:
    int removeDuplicates(int A[], int n) {
        if (n <= 2)
            return n;
        int slow = 0;
        int fast = slow + 1;
        while (fast < n) {
            // keep two copies
            if (A[fast] == A[slow] ) {

## remove_duplicates_in_sorted_array.cpp
class Solution {
public:
    int removeDuplicates(int A[], int n) {
        if (n <= 0)   //necessary, since slow is baed on 0 as in next
            return n;
        //two pointers, one slow, one fast
        // sweep fast if A[fast ] == A[slow]
        //      if new element found, keep it, set as new base element
        int moveToLeft = 0;
        int slow = 0;

## remove_element.cpp
class Solution {
public:
    int removeElement(int A[], int n, int elem) {
      //scan from beginning,
      //    if find an instance of value, copy an element from tail, and shorten by one
      //  Note: the new element from tail can NOT be equal to value.
        int tail = n-1;
        for(int i=0; i<=tail; i++) {
            if (A[i] == elem) {
                //select one non-element number

## subset_II.cpp
class Solution {
public:
/*
 Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

    Elements in a subset must be in non-descending order.
    The solution set must not contain duplicate subsets.

## binary_tree_zigzag_level_order_traversal.cpp
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {

## balanced_binary_tree.cpp
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 * Given a binary tree, determine if it is height-balanced.

## flatten_binary_tree_to_linked_list.cpp
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}\ Given a binary tree, flatten it to a linked list in-place.

For example,
Given

## symmetric_tree.cpp
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 * Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

## max_depth_of_binary_tree.cpp
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 *
 * The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

## binary_tree_level_order_traversal_II.cpp
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 * Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
	class Solution {
	public:
	int removeDuplicates(int A[], int n) {
	if (n <= 2)
	return n;
	int slow = 0;
	int fast = slow + 1;
	while (fast < n) {
	// keep two copies
	if (A[fast] == A[slow] ) {
	class Solution {
	public:
	int removeElement(int A[], int n, int elem) {
	//scan from beginning,
	// if find an instance of value, copy an element from tail, and shorten by one
	// Note: the new element from tail can NOT be equal to value.
	int tail = n-1;
	for(int i=0; i<=tail; i++) {
	if (A[i] == elem) {
	//select one non-element number
	class Solution {
	public:
	/*
	Given a collection of integers that might contain duplicates, S, return all possible subsets.

	Note:

	Elements in a subset must be in non-descending order.
	The solution set must not contain duplicate subsets.
	/**
	* Definition for binary tree
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };
	*/
	class Solution {