求常用算法的java版 拜谢了 解决方案 » 免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货 public class Sort { public static void swap(int[] data, int i, int j) { int tmp = 0; tmp = data[i]; data[i] = data[j]; data[j] = tmp; } // 插入排序 public static void sort1(int[] data) { for (int i = 1; i < data.length; i++) { for (int j = i; (j > 0) && (data[j] < data[j - 1]); j--) { swap(data, j, j - 1); } } } // 冒泡排序 public static void sort2(int[] data) { for (int i = 0; i < data.length; i++) { for (int j = data.length - 1; j > i; j--) { if (data[j] < data[j - 1]) { swap(data, j, j - 1); } } } } // 选择排序 public static void sort3(int[] data) { for (int i = 0; i < data.length; i++) { int lowIndex = i; for (int j = data.length - 1; j > i; j--) { if (data[j] < data[lowIndex]) { lowIndex = j; } } swap(data, i, lowIndex); } } // Shell排序 public static void sort4(int[] data) { for (int i = data.length / 2; i > 2; i /= 2) { for (int j = 0; j < i; j++) { for (int k = i; (k > 0) && (data[k] < data[k - 1]); k -= i) { swap(data, j, j - 1); } } } } // 快速排序 public static void sort5(int[] data) { quickSort(data, 0, data.length - 1); } private static void quickSort(int[] data, int i, int j) { int pivotIndex = (i + j) / 2; swap(data, pivotIndex, j); int k = partition(data, i - 1, j, data[j]); swap(data, k, j); if ((k - i) > 1) quickSort(data, i, k - 1); if ((j - k) > 1) quickSort(data, k + 1, j); } private static int partition(int[] data, int l, int r, int pivot) { do { while (data[++l] < pivot) ; while ((r != 0) && data[--r] > pivot) ; swap(data, l, r); } while (l < r); swap(data, l, r); return l; } // 归并排序 public static void sort6(int[] data) { int[] temp = new int[data.length]; mergeSort(data, temp, 0, data.length - 1); } private static void mergeSort(int[] data, int[] temp, int l, int r) { int mid = (l + r) / 2; if (l == r) return; mergeSort(data, temp, l, mid); mergeSort(data, temp, mid + 1, r); for (int i = l; i <= r; i++) { temp[i] = data[i]; } int i1 = l; int i2 = mid + 1; for (int cur = l; cur <= r; cur++) { if (i1 == mid + 1) data[cur] = temp[i2++]; else if (i2 > r) data[cur] = temp[i1++]; else if (temp[i1] < temp[i2]) data[cur] = temp[i1++]; else data[cur] = temp[i2++]; } } // 堆排序 public static void sort7(int[] data) { MaxHeap h = new MaxHeap(); h.init(data); for (int i = 0; i < data.length; i++) h.remove(); System.arraycopy(h.queue, 1, data, 0, data.length); } private static class MaxHeap { void init(int[] data) { this.queue = new int[data.length + 1]; for (int i = 0; i < data.length; i++) { queue[++size] = data[i]; fixUp(size); } } private int size = 0; private int[] queue; public int get() { return queue[1]; } public void remove() { swap(queue, 1, size--); fixDown(1); } // fixdown private void fixDown(int k) { int j; while ((j = k << 1) <= size) { if (j < size && queue[j] < queue[j + 1]) j++; if (queue[k] > queue[j]) // 不用交 break; swap(queue, j, k); k = j; } } private void fixUp(int k) { while (k > 1) { int j = k >> 1; if (queue[j] > queue[k]) break; swap(queue, j, k); k = j; } } }} 我也来一个,是要这个吗?public class RootOfAnEquation { /** Creates a new instance of RootOfAnEquation */ public RootOfAnEquation() { } //计算函数的值 static double Calculate(double x){ double x1 = 3*x*x*x - 5*x + 4; // 此处为待解方程的表达式 return x1; } //计算x处的导数,增量h static double diff(double x, double h){ return (Calculate(x+h)-Calculate(x))/h; } public static void main(String[] args){ double start = -5; //起始的x值 double interval = 0.0001; // 计算导数时使用的x的增量, double xn=start,xn_1=xn+1; int i=1; System.out.println("次数 x y"); //System.out.println(i + " --- " + xn + " --- " + Calculate(xn)); while (Math.abs(xn-xn_1)>0.0000001){ xn = xn_1; xn_1 = xn - Calculate(xn)/diff(xn,interval); // 牛顿迭代公式 System.out.println(i + " --- " + xn + " --- " + Calculate(xn)); i++; if (i>10000) { System.out.println("Sorry, I can't found any root of that equation for 10000 times."); break; } } }} 关于对大文件中修改个别字段的问题 c3p0 与 access 问题 java数据库大量添加数据性能问题 讨论,下面的一段代码会产生死锁吗? 【高手】 单态模式为什么不用静态类实现? 关于一个静态变量的问题 如何调用别人提供的dll 圣诞礼物,再次散分,接上次调查!圣诞绝对结帖!分不够另在加! 几个有关邮件收发难题,请教各位高手! 关于TreeSet的问题 JAVA如何自定义数据类型。 解压缩的Jar文件如何以工程形式打开
int tmp = 0;
tmp = data[i];
data[i] = data[j];
data[j] = tmp;
} // 插入排序
public static void sort1(int[] data) {
for (int i = 1; i < data.length; i++) {
for (int j = i; (j > 0) && (data[j] < data[j - 1]); j--) {
swap(data, j, j - 1);
}
}
} // 冒泡排序
public static void sort2(int[] data) {
for (int i = 0; i < data.length; i++) {
for (int j = data.length - 1; j > i; j--) {
if (data[j] < data[j - 1]) {
swap(data, j, j - 1);
}
}
}
} // 选择排序
public static void sort3(int[] data) {
for (int i = 0; i < data.length; i++) {
int lowIndex = i;
for (int j = data.length - 1; j > i; j--) {
if (data[j] < data[lowIndex]) {
lowIndex = j;
}
}
swap(data, i, lowIndex);
}
} // Shell排序
public static void sort4(int[] data) {
for (int i = data.length / 2; i > 2; i /= 2) {
for (int j = 0; j < i; j++) {
for (int k = i; (k > 0) && (data[k] < data[k - 1]); k -= i) {
swap(data, j, j - 1);
}
}
} } // 快速排序
public static void sort5(int[] data) {
quickSort(data, 0, data.length - 1);
} private static void quickSort(int[] data, int i, int j) {
int pivotIndex = (i + j) / 2;
swap(data, pivotIndex, j); int k = partition(data, i - 1, j, data[j]);
swap(data, k, j);
if ((k - i) > 1)
quickSort(data, i, k - 1);
if ((j - k) > 1)
quickSort(data, k + 1, j); } private static int partition(int[] data, int l, int r, int pivot) {
do {
while (data[++l] < pivot)
;
while ((r != 0) && data[--r] > pivot)
;
swap(data, l, r);
} while (l < r);
swap(data, l, r);
return l;
} // 归并排序
public static void sort6(int[] data) {
int[] temp = new int[data.length];
mergeSort(data, temp, 0, data.length - 1);
} private static void mergeSort(int[] data, int[] temp, int l, int r) {
int mid = (l + r) / 2;
if (l == r)
return;
mergeSort(data, temp, l, mid);
mergeSort(data, temp, mid + 1, r);
for (int i = l; i <= r; i++) {
temp[i] = data[i];
}
int i1 = l;
int i2 = mid + 1;
for (int cur = l; cur <= r; cur++) {
if (i1 == mid + 1)
data[cur] = temp[i2++];
else if (i2 > r)
data[cur] = temp[i1++];
else if (temp[i1] < temp[i2])
data[cur] = temp[i1++];
else
data[cur] = temp[i2++];
}
} // 堆排序
public static void sort7(int[] data) {
MaxHeap h = new MaxHeap();
h.init(data);
for (int i = 0; i < data.length; i++)
h.remove();
System.arraycopy(h.queue, 1, data, 0, data.length);
} private static class MaxHeap { void init(int[] data) {
this.queue = new int[data.length + 1];
for (int i = 0; i < data.length; i++) {
queue[++size] = data[i];
fixUp(size);
}
} private int size = 0; private int[] queue; public int get() {
return queue[1];
} public void remove() {
swap(queue, 1, size--);
fixDown(1);
} // fixdown
private void fixDown(int k) {
int j;
while ((j = k << 1) <= size) {
if (j < size && queue[j] < queue[j + 1])
j++;
if (queue[k] > queue[j]) // 不用交
break;
swap(queue, j, k);
k = j;
}
} private void fixUp(int k) {
while (k > 1) {
int j = k >> 1;
if (queue[j] > queue[k])
break;
swap(queue, j, k);
k = j;
}
} }}
/** Creates a new instance of RootOfAnEquation */
public RootOfAnEquation() {
}
//计算函数的值
static double Calculate(double x){
double x1 = 3*x*x*x - 5*x + 4; // 此处为待解方程的表达式
return x1;
}
//计算x处的导数,增量h
static double diff(double x, double h){
return (Calculate(x+h)-Calculate(x))/h;
}
public static void main(String[] args){
double start = -5; //起始的x值
double interval = 0.0001; // 计算导数时使用的x的增量,
double xn=start,xn_1=xn+1;
int i=1;
System.out.println("次数 x y");
//System.out.println(i + " --- " + xn + " --- " + Calculate(xn));
while (Math.abs(xn-xn_1)>0.0000001){
xn = xn_1;
xn_1 = xn - Calculate(xn)/diff(xn,interval); // 牛顿迭代公式
System.out.println(i + " --- " + xn + " --- " + Calculate(xn));
i++;
if (i>10000) {
System.out.println("Sorry, I can't found any root of that equation for 10000 times.");
break;
}
}
}
}