100以内所有可以加为100的整数的组合
import java.util.LinkedList;
public class T {
public static void main(String[] args) {
split(10, 0);
}
static LinkedList<Integer> list = new LinkedList<Integer>();
public static void split(int n, int base) {
if (n == 0) {
System.out.println(list);
return;
}
for (int i = base + 1; i <= n; i++) {
list.addLast(i);
split(n - i, i);
list.removeLast();
}
}
}
2个有序列表，将重复的部分合并为一个有序列表
public static void test() {
int[] a = { 2, 3, 4, 4, 4, 4, 7, 8, 8, 8, 8, 9, 100, 130, 150, 160 };
int[] b = { 4, 6, 7, 7, 7, 7, 8, 8, 9, 10, 100, 130, 130, 140, 150 };
int i = 0, j = 0;
ArrayList<Integer> al = new ArrayList<Integer>();
while (i < a.length && (j < b.length)) {
if (a[i] < b[j])
i++;
else if (a[i] > b[j])
j++;
else {
al.add(a[i]);
i++;
j++;
}
}
System.out.println(al);
}
任意数字字母的全排列组合问题
import java.util.Set;
import java.util.TreeSet;
public class T {
public static void perm(char[] n, int beg, int end) {
if (beg == end) {
addNumber(String.valueOf(n));
} else {
for (int i = beg; i <= end; ++i) {
swap(n, beg, i);
perm(n, beg + 1, end);
swap(n, beg, i);
}
}
}
public static void swap(char[] n, int x, int y) {
if (x == y || n[x] == n[y]) {
return;
}
char temp = n[x];
n[x] = n[y];
n[y] = temp;
}
public static void addNumber(String str) {
set.add(str);
}
public static Set<String> set = new TreeSet<String>();
public static void main(String args[]) {
char[] number = new char[] { '1', '2', '3', '4', '5' };
perm(number, 0, number.length - 1);
System.out.println(set.size());
int cols = 10;
for (String s : set) {
System.out.print(s + " ");
if (cols-- == 1) {
System.out.println();
cols = 10;
}
}
}
}
背包问题算法的JAVA实现
import java.util.Arrays;
public class T {
public static void knapsack(int[] value, int[] weight, int capicity, int[][] m) {
int n = value.length - 1;
int jMax = Math.min(weight[n] - 1, capicity);
for (int j = 0; j <= jMax; j++)
m[n][j] = 0; // 当w[n]>j 有 m[n][j]=0
showArray(m);
// 后半部分设置为此物品的价值
// m[n][j] 表示只有w[n]物品，背包的容量为j时的最大价值
for (int j = weight[n]; j <= capicity; j++)
m[n][j] = value[n]; // 当w[n]<=j 有m[n][j]=v[n]
showArray(m);
// 递规调用求出m[][]其它值，直到求出m[0][c]
for (int i = n - 1; i >= 1; i--) {
jMax = Math.min(weight[i] - 1, capicity);
System.out.println(jMax);
for (int k = 0; k <= jMax; k++)
m[i][k] = m[i + 1][k];
showArray(m);
for (int h = weight[i]; h <= capicity; h++) {
System.out.println(m[i+1][h]+" / "+ m[i + 1][h - weight[i]]+" + " +value[i]+"("+h+","+weight[i]+","+value[i]+")");
m[i][h] = Math.max(m[i + 1][h], m[i + 1][h - weight[i]] + value[i]);
}
showArray(m);
}
m[0][capicity] = m[1][capicity];
if (capicity >= weight[0])
m[0][capicity] = Math.max(m[0][capicity], m[1][capicity - weight[0]] + value[0]);
System.out.println("bestw =" + m[0][capicity]);
}
public static void showArray(int[][] m) {
for (int[] a : m) {
System.out.println(Arrays.toString(a));
}
System.out.println("-------------------------------------------");
}
public static void traceback(int[][] m, int[] w, int c, int[] x) {// 根据最优值求出最优解
int n = w.length - 1;
for (int i = 0; i < n; i++)
if (m[i][c] == m[i + 1][c])
x[i] = 0;
else {
x[i] = 1;
c -= w[i];
}
x[n] = (m[n][c] > 0) ? 1 : 0;
}
public static void main(String[] args) {
// 测试
int[] ww = { 8,5,4,3 };
int[] vv = { 10,7,5,4 };
int[][] mm = new int[4][13];
knapsack(vv, ww, 12, mm);
int[] xx = new int[ww.length];
traceback(mm, ww, 12, xx);
for (int i = 0; i < xx.length; i++)
System.out.println(xx[i]);
}
}
按数字首位逐列打印
import java.util.Queue;
import java.util.concurrent.LinkedBlockingQueue;
public class T {
public static void main(String[] args) {
int[] tArr = { 23, 2, 3, 234, 365, 564, 12, 13, 34, 35, 3453, 223, 6, 744354, 23131, 5657,
999, 877, 772, 877 };
// 10个队列
Queue<Integer>[] qs = new Queue[10];
for (int i = 0; i <= 9; i++) {
qs[i] = new LinkedBlockingQueue<Integer>();
}
for (int a : tArr) {
qs[String.valueOf(a).charAt(0) - '0'].add(a);
}
int maxSize = 0;
for (int i = 0; i <= 9; i++) {
if (qs[i].size() > maxSize) {
maxSize = qs[i].size();
}
}
Integer tmp;
// 开始显示
for (int row = 0; row < maxSize; row++) {
for (int i = 0; i <= 9; i++) {
tmp = qs[i].poll();
System.out.printf("%-7s", tmp == null ? "" : tmp);
}
System.out.println();
}
}
}
JAVA去掉一个已经排好序的数组的重复数字，尽量快
import java.util.Arrays;
public class Test {
public static void main(String args[]) {
int[] arr = { 1, 2, 2, 3, 3, 4, 4, 5, 5, 5, 6, 6, 7, 8, 9, 9, 10, 11, 11,
11, 12, 12, 13, 14, 14, 15 }; // 预设数据数组
int index = 1; // 保存最后一个不重复的位置
int last = arr[0];
for (int i = 1; i < arr.length; i++) {
if (arr[i] != last) {
arr[index] = arr[i];
last = arr[index];
index++;
}
}
int[] rtn = new int[index];
System.arraycopy(arr, 0, rtn, 0, index);
System.out.println(Arrays.toString(rtn));
}
}
按照字母排序
1. import java.util.Arrays;
2. import java.util.Comparator;
3. class T {
4. private static final int MASK = 0xFFDF; // 用来把小写变成大写
5. public static int compare(String o1, String o2) {
6. int length1 = o1.length();
7. int length2 = o2.length();
8. int length = length1 > length2 ? length2 : length1;
9. int c1, c2;
10. int d1, d2;
11. for (int i = 0; i < length; i++) {
12. c1 = o1.charAt(i);
13. c2 = o2.charAt(i);
14. d1 = c1 & MASK;
15. d2 = c2 & MASK;
16. if (d1 > d2) {
17. return 1;
18. } else if (d1 < d2) {
19. return -1;
20. } else {
21. if (c1 > c2) {
22. return 1;
23. } else if (c1 < c2) {
24. return -1;
25. }
26. }
27. }
28. if (length1 > length2) {
29. return 1;
30. } else if (length1 < length2) {
31. return -1;
32. }
33. return 0;
34. }
35.
36. public static void sortByPOPO(String[] args) {
37. String tmp;
38. for (int i = 0; i < args.length; i++) {
39. for (int j = i + 1; j < args.length; j++) {
40. if (compare(args[i], args[j]) > 0) {
41. tmp = args[i];
42. args[i] = args[j];
43. args[j] = tmp;
44. }
45. }
46. }
47. // [Ad, aC, Bc, During, day, Hello, little, X man]
48. System.out.println(Arrays.toString(args));
49. }
50.
51. public static void main(String[] args) {
52. String[] strs = { "Bc", "Ad", "aC", "Hello", "X man", "little", "During",
53. "day" };
54.
55. sortByPOPO(strs);
56.
57. }
58. }
用二分查找法判断任意整数在任意整数数组里面是否存在，若存在就返回它在数组中的索引位置，不存在返回-1
1. public class Test {
2. public int binarySearch(int[] dataset, int data, int beginIndex, int endIndex) {
3. int midIndex = (beginIndex + endIndex) / 2;
4. if (data < dataset[beginIndex] || data > dataset[endIndex] || beginIndex > endIndex)
5. return -1;
6. if (data < dataset[midIndex]) {
7. return binarySearch(dataset, data, beginIndex, midIndex - 1);
8. } else if (data > dataset[midIndex]) {
9. return binarySearch(dataset, data, midIndex + 1, endIndex);
10. } else {
11. return midIndex;
12. }
13. }
14. public int binarySearch(int[] dataset, int data) {
15. int beginIndex = 0;
16. int endIndex = dataset.length - 1;
17. int midIndex = -1;
18. if (data < dataset[beginIndex] || data > dataset[endIndex] || beginIndex > endIndex)
19. return -1;
20. while (beginIndex <= endIndex) {
21. midIndex = (beginIndex + endIndex) / 2;
22. if (data < dataset[midIndex]) {
23. endIndex = midIndex - 1;
24. } else if (data > dataset[midIndex]) {
25. beginIndex = midIndex + 1;
26. } else {
27. return midIndex;
28. }
29. }
30. return -1;
31. }
32.
33. public static void main(String args[]) {
34. Test t = new Test();
35. int[] ids = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
36. System.out.println(t.binarySearch(ids, 8, 0, ids.length-1));
37. System.out.println(t.binarySearch(ids, 8));
38. }
39. }
输入任意一个整数，返回它的阶乘
1. public class Test {
2. public long factorial(int n) {
3. // 递归
4. if (n == 1) {
5. return 1;
6. }
7. return n * factorial(n - 1);
8. }
9.
10. public long factorial2(int n) {
11. // 非递归
12. long multi = 1;
13. for (int i = 2; i <= n; i++) {
14. multi *= i;
15. }
16. return multi;
17. }
18.
19. public static void main(String args[]) {
20. Test t = new Test();
21. System.out.println(t.factorial(12));
22. System.out.println(t.factorial2(12));
23. }
24. }
能够判断任意一个整数是否素数
1. public class Test {
2. public boolean isPrimes(int n) {
3. for (int i = 2; i <= Math.sqrt(n); i++) {
4. if (n % i == 0) {
5. return false;
6. }
7. }
8. return true;
9. }
10.
11. public static void main(String args[]) {
12. Test t = new Test();
13. System.out.println(t.isPrimes(987));
14. }
15. }
给定一个java.util.Date对象，如何转化为”2007-3-22 20:23:22”格式的字符串
1. import java.text.SimpleDateFormat;
2. import java.util.Date;
3. public class Test {
4. public String dateToStr(java.util.Date date) {
5. SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
6. String str = sdf.format(date);
7. return str;
8. }
9.
10. public static void main(String args[]) {
11. Test t = new Test();
12. t.dateToStr(new Date());
13. }
14. }
