# Part I

## Data types

thanks for the help by simonmysun for the Collectionin in learning and practicing of data structure

the following comes the solutions and summary of data structure:

### Primitive types

• Given two 32-bit signed integers a and b, print how many bits changes when turning a to b.

in a word, we just need to caculate how many different bits between a and b.
first of all, we can use a^b, so that the 1 in result represent the different two bits.
second, use hamming weight to caculate the number of 1 which is the answer we want.
here is the programming by python:

class Solution:
def deor(self,a,b):
a ^= b
if a > 0:
return self.count(a)
else:
c = abs(a - 1)
return 32 - self.count(c)            #we must consider if a < 0

def count(self,n):
count = 0
while n != 0:
n = n & (n - 1)
count += 1
return count


otherwise there is a more NB way to get Hamming Weight here:

//types and constants used in the functions below

typedef unsigned __int64 uint64;  //assume this gives 64-bits
const uint64 m1 = 0x5555555555555555; //binary: 0101...
const uint64 m2 = 0x3333333333333333; //binary: 00110011..
const uint64 m4 = 0x0f0f0f0f0f0f0f0f; //binary:  4 zeros,  4 ones ...
const uint64 m8 = 0x00ff00ff00ff00ff; //binary:  8 zeros,  8 ones ...
const uint64 m16 = 0x0000ffff0000ffff; //binary: 16 zeros, 16 ones ...
const uint64 m32 = 0x00000000ffffffff; //binary: 32 zeros, 32 ones ...
const uint64 hff = 0xffffffffffffffff; //binary: all ones
const uint64 h01 = 0x0101010101010101; //the sum of 256 to the power of 0,1,2,3...

//This is a naive implementation, shown for comparison,
//and to help in understanding the better functions.
//It uses 24 arithmetic operations (shift, add, and).
int popcount_1(uint64 x) {
x = (x & m1 ) + ((x >>  1) & m1 ); //put count of each  2 bits into those  2 bits
x = (x & m2 ) + ((x >>  2) & m2 ); //put count of each  4 bits into those  4 bits
x = (x & m4 ) + ((x >>  4) & m4 ); //put count of each  8 bits into those  8 bits
x = (x & m8 ) + ((x >>  8) & m8 ); //put count of each 16 bits into those 16 bits
x = (x & m16) + ((x >> 16) & m16); //put count of each 32 bits into those 32 bits
x = (x & m32) + ((x >> 32) & m32); //put count of each 64 bits into those 64 bits
return x;
}

//This uses fewer arithmetic operations than any other known
//implementation on machines with slow multiplication.
//It uses 17 arithmetic operations.
int popcount_2(uint64 x) {
x -= (x >> 1) & m1;             //put count of each 2 bits into those 2 bits
x = (x & m2) + ((x >> 2) & m2); //put count of each 4 bits into those 4 bits
x = (x + (x >> 4)) & m4;        //put count of each 8 bits into those 8 bits
x += x >>  8;  //put count of each 16 bits into their lowest 8 bits
x += x >> 16;  //put count of each 32 bits into their lowest 8 bits
x += x >> 32;  //put count of each 64 bits into their lowest 8 bits
return x &0xff;
}

//This uses fewer arithmetic operations than any other known
//implementation on machines with fast multiplication.
//It uses 12 arithmetic operations, one of which is a multiply.
int popcount_3(uint64 x) {
x -= (x >> 1) & m1;             //put count of each 2 bits into those 2 bits
x = (x & m2) + ((x >> 2) & m2); //put count of each 4 bits into those 4 bits
x = (x + (x >> 4)) & m4;        //put count of each 8 bits into those 8 bits
return (x * h01)>>56;  //returns left 8 bits of x + (x<<8) + (x<<16) + (x<<24) + ...
}

• Given a number of height in inches and a number of height in centimeters, tell whether they equal each other.

it's so braindead so that i don't need to explain.
following是c++语言实现:

#include <iostream>
using namespace std;
int main() {
float in_cm;
float in_inches;
const float EPSINON = 0.00001;
cin >> in_cm >> in_inches;
in_inches = in_inches * 2.54;
if (- EPSINON<=in_cm-in_inches && in_cm-in_inches <= EPSINON) {
cout << "YES";
}
else {
cout << "NO";
}
}

• Explain ASCII code 0,9,10,13 and declare variables of them in charactor literals.
Orc Arvt Name
0 NUL Null
9 HT Horizontal Tab
10 LF Line Feed
13 CR Carriage Return
• How to print an emoji?
So the clang document says (emphasis mine):

This feature allows identifiers to contain certain Unicode characters, as specified by the active language standard;
Emoji List
EmojiCodeInstall

• Given n integers, each of them appears twice except for one, which appears exactly once. Find that single one.

We can use sorting to finish it with a time complexity O(nlogn). But there is a more NB way which use bitwise operators
for a solution that is O(n) time and O(1) space. As a common is the feature x^y^x=y^x^x=y. So in this case we do ^ for all
the elements of the set, finally got the answer which only appears once.
the following is the realisierung of c++ :

#include <stdio.h>
#include <iostream>
using namespace std;
int n, i;
int a;
int findsigular (int a[], int len) {
int res = 0;
int j = 0;
for (j=0;j<len;j++) {
res = res ^ a[j];
}
return res;
}
int main() {
cin >> n;
for (i=0;i<n;i++) {
cin >> a[i];
}
cout << findsigular(a,n);
}

• advanced: Given n integers, each of them appears three times except for one, which appears exactly once. Find that single one.

‘ones’ and ‘twos’ are initialized as 0. For every new element in array, find out the common set bits in the new element and previous value of ‘ones’. These common set bits are actually the bits that should be added to ‘twos’. So do bitwise OR of the common set bits with ‘twos’. ‘twos’ also gets some extra bits that appear third time. These extra bits are removed later.
Update ‘ones’ by doing XOR of new element with previous value of ‘ones’. There may be some bits which appear 3rd time. These extra bits are also removed later.
Both ‘ones’ and ‘twos’ contain those extra bits which appear 3rd time. Remove these extra bits by finding out common set bits in ‘ones’ and ‘twos’.
the following is the realisierung of c++ :

cpp
int getSingle(int arr[], int n)
{
int ones = 0, twos = 0 ;

int common_bit_mask;

// Let us take the example of {3, 3, 2, 3} to understand this
for( int i=0; i< n; i++ )
{
/* The expression "one & arr[i]" gives the bits that are
there in both 'ones' and new element from arr[].  We
add these bits to 'twos' using bitwise OR

Value of 'twos' will be set as 0, 3, 3 and 1 after 1st,
2nd, 3rd and 4th iterations respectively */
twos  = twos | (ones & arr[i]);

/* XOR the new bits with previous 'ones' to get all bits
appearing odd number of times

Value of 'ones' will be set as 3, 0, 2 and 3 after 1st,
2nd, 3rd and 4th iterations respectively */
ones  = ones ^ arr[i];

/* The common bits are those bits which appear third time
So these bits should not be there in both 'ones' and 'twos'.
common_bit_mask contains all these bits as 0, so that the bits can
be removed from 'ones' and 'twos'

Value of 'common_bit_mask' will be set as 00, 00, 01 and 10
after 1st, 2nd, 3rd and 4th iterations respectively */
common_bit_mask = ~(ones & twos);

/* Remove common bits (the bits that appear third time) from 'ones'

Value of 'ones' will be set as 3, 0, 0 and 2 after 1st,
2nd, 3rd and 4th iterations respectively */

/* Remove common bits (the bits that appear third time) from 'twos'

Value of 'twos' will be set as 0, 3, 1 and 0 after 1st,
2nd, 3rd and 4th itearations respectively */

// uncomment this code to see intermediate values
//printf (" %d %d n", ones, twos);
}

return ones;


}

## 

### Composite types or non-primitive type

• Given a string of a heximal number(might not be an integer), print it in decimal form.

according to ASCII :
Printable characters

Binary Oct Dec Hex Glyph
011 0000 060 48 30 0
011 0001 061 49 31 1
011 0010 062 50 32 2
011 0011 063 51 33 3
011 0100 064 52 34 4
011 0101 065 53 35 5
011 0110 066 54 36 6
011 0111 067 55 37 7
011 1000 070 56 38 8
011 1001 071 57 39 9
100 0001 101 65 41 A
100 0010 102 66 42 B
100 0011 103 67 43 C
100 0100 104 68 44 D
100 0101 105 69 45 E
100 0110 106 70 46 F
100 0111 107 71 47 G
... ... ... ... ...

so here is the code in c++ :

#include <iostream>
#include <string>
#include <math.h>
using namespace std;
int i, sum;
int main() {
string hexd;
cin >> hexd;
int n = hexd.length();
sum = 0;
for (i=n-1;i>=0;i--) {
if (hexd[i]>='0' && hexd[i] <= '9') {
sum+=(hexd[i]-48)*pow(16,n-i-1);
}
else if (hexd[i]>='A' && hexd[i]<='F') {
sum+=(hexd[i]-55)*pow(16,n-i-1);
}
}
cout << sum;
}


specially, if it might not be an integer, we just need do a tiny change.
find the index . and...

• Write a programm of encryption and decryption of Caesar ciphering.
#include <iostream>
#include <cmath>
#include <string.h>
using namespace std;
int main() {
char a,ch;
int k, i, j;
cout << "please enter Key(k):";
cin >> k;
cout << "please select encryption(0) or decryption(1): ";
cin >> ch;
if (ch=='0') {
cout << "please enter Plaintext string:";
cin >> a;
for (i=0; i<strlen(a); i++) {
if (a[i]>='a' && a[i]<='z') {
cout << (char)((a[i]-'a'+k)%26 + 'a');
}
else if (a[i]>='A' && a[i]<='Z') {
cout << (char)((a[i]-'A'+k)%26 + 'A');
}
}
}
else if (ch=='1') {
cout << "please enter cyphertext string: ";
cin >> a;
for (j=0;j<strlen(a);j++) {
if (a[j]>='a' && a[j]<='z') {
cout << (char)((a[j]-'a'+26-k)%26 + 'a');
}
else if (a[j]>='A' && a[j]<='Z') {
cout << (char)((a[j]-'A'+26-k)%26+'A');
}
}
}
}

• Name algorithms of string searching and compare their advantages and disadvantages.
Algorithm Preprocessing time Matching time space
Naïve string-search algorithm none Θ(nm) none
Rabin–Karp algorithm Θ(m) average Θ(n + m) O(1)
Knuth–Morris–Pratt algorithm Θ(m) Θ(n) Θ(m)
Boyer–Moore string-search algorithm Θ(m + k) best Ω(n/m) Θ(k)
Bitap algorithm Θ(m + k) O(mn)
Two-way string-matching algorithm Θ(m) O(n+m) O(1)
• Implement a expression evaluator supporting decimal numbers(with or without seperator), + and −.

acctually if we wanna figure out like 编译原理 stuff, first step is lexical analysis, then Syntacticanalysis
here is the steps:
1. we spilt each of the data, like 3*2.3+2, to 3, *, 2.3, +, 2
2. from Infix notation to RPN, 3, 2.3, *, 2, +
3. use the feature of stack to caculate the result
and there are some tiny tipps for that:
1. 符号栈top()优先级大于当前遍历的符号的话，出栈，再入当前当前遍历的符号入栈。
2. 遇’(‘左括号，符号栈保持不变，并添加’(‘入栈。
3. 遇’)’右括号，符号栈一直出栈，直到遇到’(‘左括号。

code(c++) :

#include <iostream>
#include <stack>
#include <algorithm>
#include <vector>
using namespace std;
enum AnalysisType:unsigned char{FLOAT=0x01,OPERATOR=0x02};
struct ItemValue
{
union ValueUnion
{
float fdigit;
char symbol;
};
AnalysisType type;
ValueUnion value;
};
//简单的int加减乘除
int GetPriority(char symbol)
{
switch (symbol)
{
case '+':
case '-':
return 0;
case '*':
case '/':
return 1;
case '(':
case ')':
return 2;
}
return -1;
}

bool IsOperator(char symbol)
{
switch (symbol)
{
case '+':
case '-':
case '*':
case '/':
case '(':
case ')':
return true;
}
return false;
}
//转成vector存
void wordsAnalysis(char*pIn, vector<ItemValue>&vec)
{
int nCount = strlen(pIn);
int nIndex = 0;
while (nIndex<nCount)
{
ItemValue item = { FLOAT,0};
//处理操作符
if (IsOperator(pIn[nIndex]))//如果是操作符
{
item.type = OPERATOR;
item.value.symbol = pIn[nIndex];
vec.push_back(item);
}
//处理数字
int nTemp = nIndex;
char nBuffer = {0};//20位
while (isdigit(pIn[nTemp]) || '.'==pIn[nTemp])
{
nBuffer[nTemp - nIndex]=pIn[nTemp];
++nTemp;
}
if (nTemp != nIndex)
{
item.value.fdigit = atof(nBuffer);
vec.push_back(item);
}
//////////
if (nTemp != nIndex)
nIndex += strlen(nBuffer);
else
++nIndex;
}
}
//中缀转后缀
void midToLast(vector<ItemValue> & vecIn, vector<ItemValue> & vecOut)
{
ItemValue item{ OPERATOR ,0.0f };
stack<char> stack_symbol;
for_each(std::begin(vecIn), std::end(vecIn), [&](ItemValue &it) {
if (FLOAT == it.type )
{
vecOut.push_back(it);
}
else if (OPERATOR == it.type)
{
if (')' == it.value.symbol)
{
while (!stack_symbol.empty())
{
if (stack_symbol.top() == '(')
{
stack_symbol.pop();
return;
}
item.value.symbol = stack_symbol.top();
stack_symbol.pop();
vecOut.push_back(item);
}
}
if (!stack_symbol.empty())
{
if (GetPriority(stack_symbol.top()) - GetPriority(it.value.symbol) >= 0 && stack_symbol.top()!='(')
{
item.value.symbol = stack_symbol.top();
stack_symbol.pop();
vecOut.push_back(item);
}
}
if(')' != it.value.symbol)
stack_symbol.push(it.value.symbol);
}
});
//把栈里的操作符提出来
while (!stack_symbol.empty())
{
item.value.symbol = stack_symbol.top();
stack_symbol.pop();
vecOut.push_back(item);
}
}
//计算
float Calculate(vector<ItemValue>& vec)
{
float fResult=0.0f;
stack<ItemValue> stack_value;
for_each(std::begin(vec), std::end(vec), [&](ItemValue &it) {
if (FLOAT == it.type)
{
stack_value.push(it);
}
else if(OPERATOR == it.type)
{
ItemValue Value1, Value2 ;
Value2=stack_value.top();
stack_value.pop();

Value1=stack_value.top();
stack_value.pop();

switch (it.value.symbol)
{
case '+':
Value1.value.fdigit += Value2.value.fdigit;
break;
case '-':
Value1.value.fdigit -= Value2.value.fdigit;
break;
case '*':
Value1.value.fdigit *= Value2.value.fdigit;
break;
case '/':
Value1.value.fdigit /= Value2.value.fdigit;
break;
}
Value1.type = FLOAT;
stack_value.push(Value1);
}
});
fResult = stack_value.top().value.fdigit;
return fResult;
}

int main()
{
char buffer = {0};
vector<ItemValue> vecIn,vecOut;
cout << "请输入表达式：";
cin >> buffer;
wordsAnalysis(buffer, vecIn);//拆分数据
midToLast(vecIn,vecOut);//转到后缀表达式，vecOut内
cout << "中缀到后缀：";
for_each(std::begin(vecOut), std::end(vecOut), [&](ItemValue &it) {
FLOAT == it.type?
cout << it.value.fdigit << "  ":
cout << (char)it.value.symbol << "  ";
});
cout <<endl<<"计算结果：" << Calculate(vecOut)<<endl ;
return 0;
}


for the case which only contains Operator + and - we can write more simple and explicitly instead of using Stack
here is the code:

#include <string>
#include <iostream>
using namespace std;
float calculate(float Operand1, float Operand2, char Operator) {
float res = 0 ;
if (Operator == '+') {
res = Operand1 + Operand2;
}
else if (Operator == '-') {
res = Operand1 - Operand2;
}
return res;
}

float EE(const string& str) {
float Operand1 = 0;
float Operand2 = 0;
char Operator = 0;
for (size_t i=0, size=str.size(); i<size; ++i) {
const char& ch = str[i];
if ('0'<=ch && ch<='9') {
if (Operator == 0) {
Operand1 = Operand1*10 + ch - '0';
}
else {
Operand2 = Operand2*10 + ch - '0';
}
}
else if (ch=='+' || ch=='-') {
if (Operator == 0) {
Operator = ch;
}
else {
Operand1 = calculate(Operand1, Operand2, Operator);
Operand2 = 0;
Operator = ch;
}
}
}
Operand1 = calculate(Operand1, Operand2, Operator);
return Operand1;
}
int main() {
string str;
cin >> str;
float Res = EE(str);
cout << str << "=" << Res << endl;
}