goctave

/**************************************************
1.最简单的方法就是两层循环，复杂度O(n^2)
2.使用排序，然后找重复。复杂度为O(nlgn)+O(n)=O(nlgn)
3.直接使用快排，在排序过程中，如果发现有重复的，立即结束递归
****************************************************/

#include <stdio.h>
#include <string.h>

int repeat;

goctave

#include <stdio.h>
#include <string.h>

void exchange(char str[])
{
	int len = strlen(str);
	char *beg, *end;
	char temp;
	beg = str;
	end = str + len - 1;

goctave

/************************************
O(n^2)的算法
*************************************/
#include <stdio.h>
#include <string.h>

void unique(char str[])
{
	int len = strlen(str);
	int i, j;

goctave

#include <iostream>
#include <string>
using namespace std;

bool is_anagram(const string &str1, const string &str2)
{
	if(str1.size() != str2.size())
		return false;
	int cnt1[256] = {0};
	int cnt2[256] = {0};

goctave

#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;

void replace(char *str)
{
	int n = 0;
	int i;
	for(i = 0; str[i] != '\0'; i++)

goctave

/*
	Q: Given an image represented by an NxN matrix,
	 where each pixel in the image is 4 bytes, write 
	 a method to rotate the image by 90 degrees
	 Can you do this in place?
	 
*/

/*
	C++传递二维数组比较麻烦

goctave

/*********************************************************************

	Write an algorithm such that if an element in an MxN matrix is 0, 
	its entire row and column is set to 0

**********************************************************************/
#include <iostream>
using namespace std;

const int M = 4, N = 4;

goctave

/**********************************************************************

1.Write code to remove duplicates from an unsorted linked list 
FOLLOW UP
How would you solve this problem if a temporary buffer is not allowed?

2.Implement an algorithm to find the nth to last element of 
a singly linked list. (method: find_n())

***********************************************************************/

goctave

#include <iostream>
using namespace std;

struct node
{
	int data;
	node *next;
	node(): data(0), next(NULL){}
};

goctave

//
//  main.cpp
//  cc3_3
//
//  Created by kandia on 12-5-3.
//  Copyright (c) 2012年 kandia. All rights reserved.
//

/*******************************************************************************
 * Imagine a (literal) stack of plates   If the stack gets too high, it might