本程序的思路是开一个数组，其下标表示1到m个数，数组元素的值为1表示其下标
代表的数被选中，为0则没选中。
首先初始化，将数组前n个元素置1，表示第一个组合为前n个数。
然后从左到右扫描数组元素值的“10”组合，找到第一个“10”组合后将其变为
“01”组合，同时将其左边的所有“1”全部移动到数组的最左端。
当第一个“1”移动到数组的m-n的位置，即n个“1”全部移动到最右端时，就得
到了最后一个组合。
例如求5中选3的组合：
1 1 1 0 0 //1,2,3
1 1 0 1 0 //1,2,4
1 0 1 1 0 //1,3,4
0 1 1 1 0 //2,3,4
1 1 0 0 1 //1,2,5
1 0 1 0 1 //1,3,5
0 1 1 0 1 //2,3,5
1 0 0 1 1 //1,4,5
0 1 0 1 1 //2,4,5
0 0 1 1 1 //3,4,5
求按照上述写出Java代码，要加注释的，谢谢
并且希望给出排列组合的代码，
------解决方案--------------------
另一种实现方法参考
package com.arithmeticx;
public class Combination {
//will the number of selection
private int sel_count=3;
// 存储字符
// 要取的字符目录位置
private int arr[]=new int[sel_count];
public int n=0;
public String c[] = { "1", "2", "3","4","5","6","7","8","9"};
// 从 { "1", "2", "3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18"};取出11个数进行组合，打印出所有的可能
//j代表要选取的个数，start代表选取的开始位置
public void show(int j, int start) {
for (int i = start; i < c.length; i++) {
this.arr[sel_count-j]=i;
if (j == 1) {
print();
}
if ((j - 1) != 0)
show(j-1,i+1);
else
continue;
}
}
private void print() {
for(int k=0;k<sel_count;k++){
++n;
System.out.printf("%-5s",c[arr[k]]);
}
System.out.println();
}
// 从 { "1", "2", "3","4","5","6","7","8","9"}取出11个数进行组合，打印出所有的可能
public static void main(String[] args) {
Combination c=new Combination();
c.show(c.sel_count, 0);
System.out.println(c.n+"次数");
}
}
------解决方案--------------------
- Java code
public static void main(String[] args) { int[] a = { 1, 1, 1, 0, 0 }; int i = 0; while (true) { for (int j = 0; j < a.length; j++) System.out.print(a[j]); System.out.println(); for (; i < a.length - 1; i++) { if (a[i] == 1 && a[i + 1] == 0) { int tem = a[i]; a[i] = a[i + 1]; a[i + 1] = tem; for (int j = 0, k = 0; j < i; j++) { if (j == k && a[j] == 1) { k++; continue; } if (a[j] == 1) { a[k] = a[j]; a[j] = 0; k++; } } i = 0; break; } } if(i != 0) break; } }