cli248

// #46 Fibonacci closure

package main

import "fmt"

func fibonacci() func() int {
  x, y := 0, 1
  return func() int {
    x, y = y, x+y

cli248

/*
 * Given a sorted array with duplicate elements, find the first index where its element equals to target
 * invariant: if key is in A[], then key's index in (low, high), and low + 1 < high
 */
 
int binary_search_first_position(int *A, int n, int target) {
  int low = -1, high = n;
  while (low + 1 < high) {
    int mid = low + (high - low) / 2;
    if (A[mid] < target)

cli248

void insertNode(LinkedList *newIp, LinkedList *list) {

  LinkedList **lpp;
  LInkedList *lp;

  for (lpp = &list; *lpp != NULL; lpp = &(*lpp)->next) {
    lp = *lpp;
    if (newIp->val < lp->val) {
      newIp->next = lp;
      *lpp = newIp;

cli248

struct Node {
  Node *left;
  Node *right;
  int _data;
  Node(int data) : _data(data), left(NULL), right(NULL) {}
};

bool checkBST(Node *root) {
  return help(root, INT_MIN, INT_MAX);
}

cli248

/*
 Credit to Leetcode: http://leetcode.com/2010/04/reversing-linked-list-iteratively-and.html
 */
 
struct Node {
  int val;
  Node *next;
  Node(int value):val(value), next(NULL) {};
};

cli248

/*
 * Credits: http://stackoverflow.com/questions/8599374/sub-sequence-occurrence-in-a-string
 */

bool checkSubSequence(string A, string B) {
  int pos = -1;
  bool ok = true;
  
  for (int i=0; i!=A.size() && ok; i++)
    ok = (pos = B.find(A[i], pos+1)) != string::npos;

cli248

class Queue():
  def __init__(self):
    self.inbox = []
    self.outbox = []
    
  def enqueue(self, item):
    self.inbox.append(item)
    
  def dequeue(self):
    if not self.outbox: