算法 Robert Sedgewick 习题答案 1.1 基础编程模型

似乎Algorithms 4th没有答案，那就自己做一份吧，难免有错误

stdlib,algis4两个jar包去此书官网下载导入工程

main中为我自己写的测试程序

1.1.1

7

200.0000002

true

1.1.2 

double 1.618

double  10.0

boolaen true

String 33

1.1.3

package chapter1_1;

public class Exercise3 {

public static void main(String[] args) {
if (new Exercise3().isEqual(args[0], args[1], args[2]))
System.out.println("equal");
else
System.out.println("not equal");
}

private boolean isEqual(String s1, String S2, String S3) {
return Integer.parseInt(s1) == Integer.parseInt(S2)
&& Integer.parseInt(S2) == Integer.parseInt(S3);

}
}

1.1.4

不需要then

少括号

正确

少分号

1.1.5

package chapter1_1;

public class Exercise3 {

public static void main(String[] args) {
if (new Exercise3().isEqual(args[0], args[1], args[2]))
System.out.println("equal");
else
System.out.println("not equal");
}

private boolean isEqual(String s1, String S2, String S3) {
return Integer.parseInt(s1) == Integer.parseInt(S2)
&& Integer.parseInt(S2) == Integer.parseInt(S3);

}
}

1.1.6

0

1

1

2

3

5

8

13

21

34

55

89

144

233

377

610

斐波那契数列

1.1.7

3.00009

499500

10000

1.1.8

b

197

101

1.1.9

package chapter1_1;

public class Exercise9 {
public static void main(String[] args)
{
System.out.println(new Exercise9().toBinaryString(525));
}

//N用二进制表示并
private String toBinaryString(int N)
{
String s = "";
for(int n=N;n>0;n=n>>1)
s = (n&1) + s;
return s;
}
}

1.1.10

数组没有初始化

1.1.11

package chapter1_1;

public class Exercise11 {
public static void main(String[] args)
{
boolean[][] a={{true,true},{false,true}};
System.out.print(" 0 1");
System.out.println();
for(int i=0;i<2;i++)
{
System.out.print(i);
for(int j=0;j<2;j++)
if(a[i][j])
System.out.print("*"+" ");
else System.out.print(" "+" ");
System.out.println();
}
}
}

1.1.12

0

1

2

3

4

5

6

7

8

9

1.1.13

package chapter1_1;

public class Exercise13 {

private int[][] transposition(int[][] a, int M, int N) {
int[][] b = new int
[M];
for (int i = 0; i < N; i++)
for (int j = 0; j < M; j++)
b[i][j] = a[j][i];
return b;
}

public static void main(String[] args) {
int[][] a = { { 1, 2, 3, 4 }, { 3, 2, 3, 5 }, { 6, 3, 6, 8 } };

a = new Exercise13().transposition(a, 3, 4);

for (int i = 0; i < 4; i++) {
for (int j = 0; j < 3; j++)
System.out.print(a[i][j] + " ");
System.out.println();
}
}
}

1.14

package chapter1_1;

public class Exercise14 {
public static int lg(int N) {
int count = 0;
while (N != 1) {
N = N >> 1;
++count;

}
return count;
}

public static void main(String[] args) {
System.out.println(Exercise14.lg(33));

}
}

1.1.15

package chapter1_1;

public class Exercise15 {
public static void main(String[] args) {
int[] a = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 6, 5, 3 };
int[] b = Exercise15.histogram(a, 11);
int sum = 0;
for (int i = 0; i < b.length; i++) {
System.out.print(b[i] + " ");
sum += b[i];

}
// a[]中所有元素都在0到M-1之间，b中元素的和应该和a的长度相等
System.out.println(sum == a.length);
}

private static int[] histogram(int[] a, int M) {
int[] b = new int[M];
for (int i = 0; i < a.length; i++)
if (a[i] >= 0 && a[i] <= M - 1)
b[a[i]]++;

return b;
}
}

1.1.16

311361142246

字符串+整形的应用

1.1.17

基础情况永远不会被访问

1.1.18

50

33

a,b的积

33554432

177147

a的b次方

1.1.19

计算到40以上效率明显变慢（没有测试1个小时）

package chapter1_1;

public class Exercise19 {
// 递归实现斐波那契数列
public static long F(int N) {
if (N == 0)
return 0;
if (N == 1)
return 1;
return F(N - 1) + F(N - 2);
}

public static void main(String[] args) {
for (int N = 0; N < 100; N++)
System.out.println(N + " " + F1(N));
}

// 循环实现斐波那契数列
public static long F1(int N) {
if (N == 0)
return 0;
int[] a = new int[N + 1];
a[0] = 0;
a[1] = 1;
for (int i = 2; i < a.length; i++)
a[i] = a[i - 1] + a[i - 2];
return a
;

}
// TODO 矩阵快速幂实现斐波那契数列，此题效率已经满足要求，以后写
}

1.1.20

package chapter1_1;

public class Exercise20 {

public static double ln(int N) {

if (N == 1)
return Math.log(1);

return Math.log(N) + ln(N - 1);
}

public static void main(String[] args) {
System.out.println(ln(9));

}

}

1.1.21

使用的测试数据为2012-2013赛季西甲射手榜，输入为姓名，进球数，射门数

Messi|46|235

Ronaldo|34|163

Falcao|28|123

Negredo|25|150

Soldado|24|100

Castro|18|79

Piti|18|111

Higuaín|16|56

Aduriz|14|87

Alberto|14|102

package chapter1_1;

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.util.ArrayList;

public class Exercise21 {
public static void main(String[] args) {
try {
BufferedReader br = new BufferedReader(new FileReader("goals.txt"));

String s = null;
ArrayList<forward> forwards = new ArrayList<>();

while ((s = br.readLine()) != null) {
forward f = readData(s);
forwards.add(f);
}

System.out.println("name      " + "goals      " + "shots     "
+ "accuracy");
for (forward f : forwards)
System.out.print(f);

} catch (Exception e) {
e.printStackTrace();
}
}

public static forward readData(String s) {
String[] datas = s.split("\\|");

return new forward(datas[0], Integer.parseInt(datas[1]),
Integer.parseInt(datas[2]));

}
}

class forward {
String name;
int goals;
int shots;
double accuracy;

public forward(String name, int goals, int shots) {
this.name = name;
this.goals = goals;
this.shots = shots;
this.accuracy = (double) goals / shots;
}

@Override
public String toString() {
return name + "        " + goals + "       " + shots + "     "
+ accuracy + "\n";

}
}

1.1.22

package chapter1_1;

public class Exercise22 {

private int count;

public int rank(int key, int lo, int hi, int[] a) {
++count;
for (int i = 0; i < count; i++)
System.out.print(" ");
System.out.println("lo=" + lo + "  hi=" + hi);
if (lo > hi)
return -1;
int mid = (hi - lo) / 2 + lo;
if (key < a[mid])
return rank(key, lo, mid - 1, a);
else if (key > a[mid])
return rank(key, mid + 1, hi, a);
else
return mid;
}

public static void main(String[] args) {
int[] a = { 1, 3, 5, 6, 7, 8, 9, 11, 13, 15, 16, 17, 35, 67, 78, 79 };
System.out.println(new Exercise22().rank(16, 0, a.length - 1, a));
System.out.println(new Exercise22().rank(18, 0, a.length - 1, a));
}
}

1.1.23

使用的查找名单为（whiteList.txt）

3

2

1

237

343452

123

12

12333

756

43

23

433

233

55

554

332

221

7666

553344

45425

2524

32442

233444

5

5525

48789

87

3269

998

646

997

777777

46554544

456488994

6464646

64465

54656

4664

468

64684

898

894

99999

4653

44

48

package chapter1_1;

import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;

public class Exercise23 {

public int rank(int key, int lo, int hi, ArrayList<Integer> arry) {
int mid = (hi - lo) / 2 + lo;
if (lo > hi)
return -1;

if (key > arry.get(mid))
return rank(key, mid + 1, hi, arry);
else if (key < arry.get(mid))
return rank(key, lo, mid - 1, arry);
else
return mid;
}

public void search(ArrayList<Integer> arry, int[] search, char c) {
int[] a = new int[search.length];
for (int i = 0; i < search.length; i++)
a[i] = rank(search[i], 0, arry.size() - 1, arry);
if (c == '+') {
for (int j = 0; j < a.length; j++) {
if (a[j] == -1)
System.out.print(search[j] + " ");
}

} else {
for (int j = 0; j < a.length; j++) {
if (a[j] != -1)
System.out.print(search[j] + " ");
}
}

}

public static void main(String args[]) {
ArrayList<Integer> arry = new ArrayList<>();
try {
BufferedReader br = new BufferedReader(new FileReader(
"mydata/whiteList.txt"));
String s = null;
while ((s = br.readLine()) != null) {
arry.add(Integer.parseInt(s));
}

} catch (Exception e) {

e.printStackTrace();
}

Collections.sort(arry);

int[] search = { 2, 3, 45, 23, 77, 3269, 9787, 693, 468, 8941, 6666666,
99999, 777777 };

Exercise23 binarySearch = new Exercise23();
binarySearch.search(arry, search, '+');
System.out.println();
binarySearch.search(arry, search, '-');

}
}

1.1.24

package chapter1_1;

public class Exercise24 {

public static int Euclid(int p,int q)
{
System.out.println("p="+p+" "+"q="+q);
if(q==0)
return p;
else
return Euclid(q,p%q);
}

public static void main(String[] args)
{
System.out.println(Euclid(105,24));
System.out.println(Euclid(Integer.parseInt(args[0]), Integer.parseInt(args[1])));
}
}


p=105 q=24

p=24 q=9

p=9 q=6

p=6 q=3

p=3 q=0

3

p=1111111 q=1234567

p=1234567 q=1111111

p=1111111 q=123456

p=123456 q=7

p=7 q=4

p=4 q=3

p=3 q=1

p=1 q=0

1

1.1.25

首先，算法的最终结果rN−1同时整除a和b。因为它是一个公约数，所以必然小于或者等于最大公约数g。然后，任何a和b的公约数都能整除rN−1。所以g一定小于或等于rN−1。两个不等式只在rN−1 = g是同时成立。具体证明如下：

证明rN−1同时整除a和b：
因为余数rN是0，rN−1能够整除rN−2：
rN−2 = qNrN−1因为rN−1能够整除rN−2，所以也能够整除rN−3：
rN−3 = qN−1
rN−2 + rN−1同理可证rN−1可以整除所有之前步骤的余数，包括a和b，即rN−1是a和b的公约数，rN−1 ≤ g。

证明任何整除a和b的自然数g（也就是a和b的公约数）都能整除rN-1：
根据定义，a和b可以写成g的倍数：a = mg、b = ng，其中m和n是自然数。因为r0 = a − q0b = mg − q0ng = (m − q0n)g，所以g整除r0。同理可证g整除每个余数r1、r2……rN-1。所以，最大公约数g一定整除rN−1，因而g ≤ rN−1。

因为第一步的证明告诉我们rN−1 ≤ g，所以g = rN−1。即：

g = GCD(a, b) = GCD(b, r0) = GCD(r0,
r1) = … = GCD(rN−2, rN−1) =
rN−1