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   -> 数据结构与算法 -> 数据结构与算法JAVA语言描述第一、二章部分课后习题参考答案 -> 正文阅读

[数据结构与算法]数据结构与算法JAVA语言描述第一、二章部分课后习题参考答案

Ch1:

1.14:

public class OrderedCollection {

??? private Comparable[] comparables;

??? public Comparable[] getComparables() {

??????? return comparables;

??? }

??? public void setComparables(Comparable[] comparables) {

??????? this.comparables = comparables;

??? }

??? public boolean isEmpty() {

??????? if (comparables == null || comparables.length == 0) {

??????????? return true;

??????? }

??????? return false;

??? }

??? public void makeEmpty() {

?????? ?comparables = new Comparable[]{};

??? }

??? public void insert(Comparable comparable) {

??????? int length = this.comparables.length;

??????? Comparable[] comparables = new Comparable[length + 1];

??????? System.arraycopy(this.comparables, 0, comparables, 0, length);

??????? comparables[length] = comparable;

??????? this.comparables = comparables;

??? }

??? public void remove(int index) {

??????? if (index < 0 || this.comparables == null || this.comparables.length == 0) {

??????????? return;

??????? }

??????? int length = this.comparables.length;

??????? if (index > length - 1) {

??????????? return;

??????? }

??????? Comparable[] objects = new Comparable[length - 1];

??????? for (int i = 0; i < length; i++) {

??????????? if (i == index) {

???? ???????????continue;

??????????? }

??????????? if (i < index) {

??????????????? objects[i] = comparables[i];

??????????? } else {

??????????????? objects[i - 1] = comparables[i];

??????????? }

??????? }

??????? this.comparables = objects;

??? }

??? public Comparable findMax() {

??????? if (comparables == null || comparables.length == 0) {

??????????? return null;

??????? }

??????? int maxIndex = 0;

??????? for (int i = 0; i < comparables.length; i++) {

??????????? if (comparables[i].compareTo(comparables[maxIndex]) > 0) {

??????????????? maxIndex = i;

??????????? }

??????? }

??????? return comparables[maxIndex];

??? }

??? public Comparable findMin() {

??????? if (comparables == null || comparables.length == 0) {

??????????? return null;

??????? }

??????? int minIndex = 0;

??????? for (int i = 0; i < comparables.length; i++) {

??????????? if (comparables[i].compareTo(comparables[minIndex]) < 0) {

??????????????? minIndex = i;

??????????? }

??????? }

??????? return comparables[minIndex];

??? }

}

class Test {

??? public static void main(String[] args) {

??????? OrderedCollection collection = new OrderedCollection();

??????? collection.setComparables(new Comparable[]{1, 2, 3, 4, 5, 6, 7, 8});

??????? collection.insert(9);

??????? collection.remove(7);

??????? System.out.println(collection.findMax());

??????? System.out.println(collection.findMin());

??????? for (Comparable comparable : collection.getComparables()) {

??????????? System.out.print(comparable + " ");

??????? }

??? }

}

Ch2

2.7:

(1):

a:The running time is O(N)

b: The running time in Java is shown in the figure below

c: My analysis is basically consistent with the real running time

(2):

a:The running time is O(N2)

b: The running time in Java is shown in the figure below

c:My analysis is basically consistent with the real running time

(3):

a:The running time is O(N3)

b: The running time in Java is shown in the figure below

c: My analysis takes longer than the real run time

(4):

a:The running time is O(N2)

b: The running time in Java is shown in the figure below

c:My analysis is basically consistent with the real running time

(5):

a:The running time is O(N5)

b: The running time in Java is shown in the figure below. But this program takes too long for my computer to calculate

c:I think that my analysis takes longer than the real run time, but I didn't get the real time . So I have no facts to prove it

(6):

a:The running time is O(N4)

b: The running time in Java is shown in the figure below. But this program takes too long for my computer to calculate

c: I think that my analysis is basically consistent with the real running time. But I didn't get the real time. So I have no facts to prove it

2.11:

a:It takes 2.5ms(0.5ms * 5 = 2.5 ms)

b:It takes slightly more than 2.5ms(0.5ms * 5log5 >2.5ms)

c:It takes 12.5ms(0.5ms * 25 = 12.5ms)

d:It takes 62.5ms(0.5ms * 125 = 62.5ms)

2.17:

a:

public class Demo7 {

??? public static void main(String[] args) {

??? ????int[] arr = {0, 1, 2, -5, -7, 6, 8};

??????? int minSeq = getMinSeq(arr);

??????? System.out.println(minSeq);

??? }

??? public static int getMinSeq(int[] arr) {

??????? int minSum = 0;

??????? if (arr != null && arr.length > 0) {

??????????? minSum = arr[0];

??????????? int thisSum = 0;

??????????? for (int i = 0; i < arr.length; i++) {

??????????????? thisSum += arr[i];

??????????????? if (thisSum < minSum) {

??????????????????? minSum = thisSum;

??????????????? } else if (thisSum > 0) {

??????????????????? thisSum = 0;

??????????????? }

??????????? }

??????? }

??????? return minSum;

??? }

}

Running time analyses: The running time is O(N)

b:

public class Demo8 {

??? public static void main(String[] args) {

??????? int[] arr = {2, -3, 8, -6, 12, 9, 5, -2};

??????? int minPositiveSeq = getMinPositiveSeq(arr);

??????? System.out.println(minPositiveSeq);

??? }

??? public static int getMinPositiveSeq(int[] arr) {

??????? int minPosSum = 0;

??????? int len = arr.length;

????? ??int[] newArr = new int[len];

??????? for (int i = 0; i < len; i++) {

??????????? minPosSum += arr[i];

??????????? newArr[i] = minPosSum;

??????? }

??????? int[] newArray = new int[newArr.length];

??????? System.arraycopy(newArr, 0, newArray, 0, newArr.length);

??????? quickSort(newArr);

??????? int min = newArr[0] >= 0 ? newArr[0] : newArr[len - 1];

??????? for (int i = 1; i < len; i++) {

??????????? if (newArr[i] > newArr[i - 1]) {

??????????????? int temp = newArr[i] - newArr[i - 1];

??????????????? if (temp < min) {

??????????????????? int i1 = search(newArray, newArr[i]);

??????????????????? int i2 = search(newArray, newArray[i - 1]);

??????????????????? if (i1 > i2) {

??????????????????????? min = temp;

??????????????????? }

??????????????? }

??????????? }

??????? }

??????? return min;

??? }

??? public static int search(int[] arr, int key) {

??????? for (int i = 0; i < arr.length; i++) {

??????????? if (arr[i] == key) {

??????????????? return i;

??????????? }

??????? }

??????? return 0;

??? }

??? public static void quickSort(int[] array) {

??????? if (array == null || array.length == 0 || array.length == 1) {

??????????? return;

??????? }

??????? sort(array, 0, array.length - 1);

??? }

??? public static void sort(int[] array, int left, int right) {

??????? if (left > right) {

??????????? return;

??????? }

??????? int base = array[left];

??????? int i = left, j = right;

??????? while (i != j) {

??????????? while (array[j] >= base && i < j) {

??????????????? j--;

??????????? }

??????????? while (array[i] <= base && i < j) {

??????????????? i++;

??????????? }

??????????? if (i < j) {

??????????????? int temp = array[i];

??????????????? array[i] = array[j];

??????????????? array[j] = temp;

??????????? }

??????? }

??????? array[left] = array[i];

??????? array[i] = base;

??????? sort(array, left, i - 1);

??????? sort(array, i + 1, right);

??? }

}

Running time analyses: The running time is O(NlogN)

C:

public class Demo9 {

??? public static void main(String[] args) {

??????? int [] arr = {3,5,-4,8,7,-6,9};

??????? int maxSeqMulti = getMaxSeqMulti(arr);

??????? System.out.println(maxSeqMulti);

??? }

??? public static int getMaxSeqMulti(int [] arr){

??????? int max = Integer.MIN_VALUE;

??????? for(int i = 0;i<arr.length;i++){

??????????? int temp = arr[i];

??????????? for (int j = i+1;j<arr.length;j++){

??????????????? max = Math.max(max,temp);

??????????????? temp*=arr[j];

??????????? }

??????????? max = Math.max(max,temp);

??????? }

??????? return max;

??? }

}

Running time analyses: The running time is O(N2)

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教程: HTML教程 CSS教程 JavaScript教程 Go语言教程 JQuery教程 VUE教程 VUE3教程 Bootstrap教程 SQL数据库教程 C语言教程 C++教程 Java教程 Python教程 Python3教程 C#教程
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