给出一串数字，算出所有组合值
public class AllOrderWithinNumbers {public static void main(String...args){int[] i = new int[]{1,2,4,5};AllOrderWithinNumbers t = new AllOrderWithinNumbers();ArrayList al = t.getOrders(i);System.out.println(al.size());System.out.println(al);}private ArrayList getOrders(int[] others){ArrayList al = new ArrayList();//如果只有一个数，则返回自身if(others.length==1){al.add(others[0]);}else{for(int i=0;i<others.length;i++){int[] t = this.getLeftNumber(others,i);ArrayList tmp = this.getOrders(t);//tmp.add(0,others[i]); //uncomment this line and comment the blow line, to see what the result isthis.linkSubOrders(others[i],tmp);//这里要进行判断，如果tmp中包含ArrayList，那么直接addAll，把每一个ArrayList作为子元件插入到al中，如果tmp不包含ArrayList，那么要把tmp作为一个整体插入//al.add(tmp);// this line and the next 4 lines, you can compare the differencesif(tmp.get(0) instanceof ArrayList){al.addAll(tmp);}else{al.add(tmp);}}}return al;}/** * * @param i 开头数 * @param tmp 除了开头数之外其它数字组成的排序，由于在多个ArrayList中，因此要把开头数插入到每个ArrayList开头 * 注意：由于tmp中会存在嵌套的ArrayList，因此这里要进行嵌套ArrayList递归，如[1,[2,[3,4][4,3]],[3,[2,4][4,2]]...]要转换成[1,2,3,4],[1,2,4,3],[1,3,2,4]...这样的形式 */private void linkSubOrders(int i, ArrayList tmp) {if(tmp.get(0) instanceof ArrayList){for(int j=0;j<tmp.size();j++){ArrayList al = (ArrayList) tmp.get(j);this.linkSubOrders(i, al);}}else{tmp.add(0,i);}}/** * copy others to a new array, not including the i-th number * @param others * @param i * @return */private int[] getLeftNumber(int[] others, int i) {int[] t = new int[others.length-1];for(int j=0;j<others.length;j++){if(j<i){t[j] = others[j];}if(j==i){}if(j>i){t[j-1] = others[j];}}return t;}}private void test(){ char[] dictionaries = "ABC".toCharArray(); int limit = 1 << dictionaries.length; for (int mask = 1; mask < limit; mask++) { for (int submask = 1, i = 0; i < dictionaries.length; submask <<= 1, i++) { if ((mask & submask) != 0) { System.out.print(dictionaries[i]); } } System.out.println(); } }