fever324

/*
Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1
   / \
  2   2
 / \ / \

fever324

import java.util.Stack;

/*
 * Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
 *
 * push(x) -- Push element x onto stack.
 * pop() -- Removes the element on top of the stack.
 * top() -- Get the top element.
 * getMin() -- Retrieve the minimum element in the stack.
 */

fever324

import java.util.LinkedList;
import java.util.Stack;

/*
 * Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
 *
 * push(x) -- Push element x onto stack.
 * pop() -- Removes the element on top of the stack.
 * top() -- Get the top element.
 * getMin() -- Retrieve the minimum element in the stack.

fever324

/*
 * Given a list, rotate the list to the right by k places, where k is non-negative.
 *
 * For example:
 * Given 1->2->3->4->5->NULL and k = 2,
 * return 4->5->1->2->3->NULL.
 */
public class Solution {
    public ListNode rotateRight(ListNode head, int n) {
        // make sure list length > 1

fever324

public class Solution1 {
    public int reverse(int x) {

        if(x < 10 && x > -10){
            return x;
        }
        boolean negative = x < 0;
        char[] c = null;
        if(negative) {
            c = (x+"").substring(1).toCharArray();

fever324

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }

fever324

/*
Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character
*/

fever324

/*
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).

For example:
Given binary tree {3,9,20,#,#,15,7},
    3
   / \
  9  20
    /  \
   15   7

fever324

/*
Given two binary strings, return their sum (also a binary string).

For example,
a = "11"
b = "1"
Return "100".
*/

public class Solution {

fever324

/*
Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.
*/

public class Solution {
    public int majorityElement(int[] num) {
        int candidate = num[0];
        int counter = 0;