数据结构实验一顺序表问题及实验报告模板 - Copy

for (int t = 0; t < testcases.length; t++) {
String target = testcases[t];
System.out.println(target
+ " " + linearSearch1(word1, left1, right1, target) + " "
}
public static int linearSearch2(Comparable[] a, int left, int right, Comparable val)
{
for (int i = left; i <= right; i++)
{ int comp = val.compareTo(a[i]);
if (comp == 0)
return i;
else if (comp < 0)
break;
}
return NONE;
}
public static int binarySearch(Comparable[] a,
int left, int right, Comparable val)
{
int l = left, r = right;
while (l <= r) {
int m = (l + r) / 2;
int comp = val.compareTo(a[m]);
if (comp == 0) return m;
else if (comp < 0) r = m - 1;
else l = m + 1;
}
return NONE;
}源自文库
public static void main(String[] args) {
{
for (int i = left; i <= right; i++)
{ int comp = val.compareTo(a[i]);
if (comp == 0)
return i;
else if (comp < 0)
break;
}
return NONE;
}
public static int binarySearch(Comparable[] a,
+ linearSearch2(word2, left2, right2, target)
+ " " + binarySearch(word2, left2, right2, target));
}
}
}
三、运行输出结果：
四、调试和运行程序过程中产生的问题及采取的措施：
2.从键盘输入1个整数，在顺序表中查找该结点。若找到，输出结点的位置；若找不到，则显示“找不到”。
调试数据：第一次输入11，第二次输入3
3.从键盘输入2个整数，一个表示欲插入的位置i，另一个表示欲插入的数值x，将x插入在对应位置上，输出顺序表所有结点值，观察输出结果。
调试数据：第一次insert "11" after "6"，第二次insert"86"at"2"
int left1 = 0, right1 = word1.length - 1;
String[] word2 = {"cat", "dog", "eel", "fox", "pig", "rat"};
int left2 = 0, right2 = word2.length - 1;
String[] testcases = {"ant", "bat", "cat", "dog", "eel","fox", "hen", "pig", "rat"};
+ " " + binarySearch(word2, left2, right2, target));
}
}
}
程序二
public class Searching {
private static final int NONE = -1;
public static int linearSearch1(Object[] a,int left, int right, Object val)
第四题
delete the number at "2":head -> 9 -> 8-> 6 -> 5 -> 4 -> 3 -> 2 -> 1 ->null
delete value "9": head -> 8 -> 7 -> 6 -> 5-> 4 -> 3 -> 2 -> 1-> null
二、源程序及注释：
第二题
找不到
6
第三题
insert "11" after "6": head -> 9 -> 8 -> 7 -> 6 ->11-> 5 -> 4 -> 3 -> 2 -> 1 -> null
insert"86"at"2":head -> 9 -> 8 -> 86 -> 7 -> 6 -> 5 -> 4 -> 3 -> 2 -> 1 ->null
{
for (int i = left; i <= right; i++)
if (val.equals(a[i]))
return i;
return NONE;
}
public static int linearSearch2(Comparable[] a, int left, int right, Comparable val)
int left2 = 0, right2 = word2.length - 1;
String[] testcases = {"ant", "bat", "cat", "dog", "eel","fox", "hen", "pig", "rat"};
for (int t = 0; t < testcases.length; t++) {
else if (comp < 0) r = m - 1;
else l = m + 1;
}
return NONE;
}
public static void main(String[] args) {
String[] word1 = {"dog", "cat", "rat", "pig", "fox", "eel"};
String target = testcases[t];
System.out.println(target
+ " " + linearSearch1(word1, left1, right1, target) + " "
+ linearSearch2(word2, left2, right2, target)
以下程序仅供参考，其结果不代表本实验要求。学生可输入以下程序，调试并运行之，分析输出结果，然后结合书本上的算法进行修改，完成上述实验内容，请实验完成之后，把正确的程序复制在此处。程序一：
public class Searching {
private static final int NONE = -1;
实验一顺序表问题
【实验报告】
《数据结构与算法》实验报告一
学院：计算机与信息学院
班级：
学号：
姓名：
日期：
程序名：
一、上机实验的问题和要求：
顺序表的查找、插入与删除。设计算法，实现线性结构上的顺序表的产生以及元素的查找、插入与删除。具体实现要求：
1.从键盘输入10个整数，产生顺序表，并输出结点值。
调试数据：9 8 7 6 5 4 3 2 1
4.从键盘输入1个整数，表示欲删除结点的位置，输出顺序表所有结点值，观察输出结果。
调试数据：第一次delete the number at "2"，第二次delete value "9"
注意：顺序表输出表现形式如下（实验报告上为截图）：
顺序表：
第一题
Initially Seq: head -> 9 -> 8 -> 7 -> 6 -> 5 -> 4 -> 3 -> 2 -> 1 -> null
String[] word1 = {"dog", "cat", "rat", "pig", "fox", "eel"};
int left1 = 0, right1 = word1.length - 1;
String[] word2 = {"cat", "dog", "eel", "fox", "pig", "rat"};
int left, int right, Comparable val)
{
int l = left, r = right;
while (l <= r) {
int m = (l + r) / 2;
int comp = val.compareTo(a[m]);
if (comp == 0) return m;
public static int linearSearch1(Object[] a,int left, int right, Object val)
{
for (int i = left; i <= right; i++)
if (val.equals(a[i]))
return i;
return NONE;