/**
* 普通数组的Java代码
* @author dream
*
*/
public class GeneralArray {
private int[] a;
private int size; //数组的大小
private int nElem; //数组中有多少项
public GeneralArray(int max){
this.a = new int[max];
this.size = max;
this.nElem = 0;
}
public boolean find(int searchNum){ //查找某个值
int j;
for(j = 0; j < nElem; j++){
if(a[j] == searchNum){
break;
}
}
if(j == nElem){
return false;
}else {
return true;
}
}
public boolean insert(int value){ //插入某个值
if(nElem == size){
System.out.println("数组已满");
return false;
}
a[nElem] = value;
nElem++;
return true;
}
public boolean delete(int value){ //删除某个值
int j;
for(j = 0; j < nElem; j++){
if(a[j] == value){
break;
}
}
if(j == nElem){
return false;
}
if(nElem == size){
for(int k = j; k < nElem - 1; k++){
a[k] = a[k+1];
}
}else {
for(int k = j; k < nElem; k++){
a[k] = a[k+1];
}
}
nElem--;
return true;
}
public void display(){ //打印整个数组
for(int i = 0; i < nElem; i++){
System.out.println(a[i] + " ");
}
System.out.println("");
}
}
/**
* 有序数组的Java代码
* @author dream
*
*/
/**
* 对于数组这种数据结构,
* 线性查找的话,时间复杂度为O(N),
* 二分查找的话时间为O(longN),
* 无序数组插入的时间复杂度为O(1),
* 有序数组插入的时间复杂度为O(N),
* 删除操作的时间复杂度均为O(N)。
* @author dream
*
*/
public class OrderedArray {
private long[] a;
private int size; //数组的大小
private int nElem; //数组中有多少项
public OrderedArray(int max){ //初始化数组
this.a = new long[max];
this.size = max;
this.nElem = 0;
}
public int size(){ //返回数组实际有多少值
return this.nElem;
}
/**
* 二分查找
* @param searchNum
* @return
*/
public int find(long searchNum){
int lower = 0;
int upper = nElem - 1;
int curr;
while (true) {
curr = (lower + upper) / 2;
if(a[curr] == searchNum){
return curr;
}else if(lower > upper){
return -1;
}else {
if(a[curr] < searchNum){
lower = curr + 1;
}else {
upper = curr - 1;
}
}
}
}
public boolean insert(long value){ //插入某个值
if(nElem == size){
System.out.println("数组已满!");
return false;
}
int j;
for(j = 0; j < nElem; j++){
if(a[j] > value){
break;
}
}
for(int k = nElem; k > j; k++){
a[k] = a[k-1];
}
a[j] = value;
nElem++;
return true;
}
public boolean delete(long value){ //删除某个值
int j = find(value);
if(j == -1){
System.out.println("没有该元素!");
return false;
}
if(nElem == size){
for(int k = j; k < nElem - 1; k++){
a[k] = a[k+1];
}
a[nElem-1] = 0;
}else {
for(int k = j; k < nElem; k++){
a[k] = a[k+1];
}
}
nElem--;
return true;
}
public void display(){ //打印整个数组
for(int i = 0; i < nElem; i++){
System.out.println(a[i] + " ");
}
System.out.println("");
}
}