Speak up is relatively simple, from 0 to N traverse the output on the line, but if N is very large, involving integer overflow issue, it is clear that a full permutation problem, which is the output of N, on behalf of all the digital values of the N-bit is 0 -9 make a full array, also we need to consider this is similar to the front 0001,0068,0977 0 when the output needs to remove.
Is a recursive basis of perfection permutation problem, of course, can not use recursion stack to solve, here is a recursive version of java.
public class Test {
public static void recur(int[] arr, int index){
if(index > arr.length - 1){
// 输出
//判断从第几位开始输出
int start_id = 0;
for(int j = 0; j < arr.length; j ++){
if(arr[j] == 0){
start_id += 1;
}
else{
break;
}
}
for(int j = start_id; j < arr.length; j ++){
System.out.print(arr[j]);
}
System.out.println();
return;
}
for(int i = 0; i <= 9; i ++){
arr[index] = i;
recur(arr, index + 1);
}
arr[index] = 0;
}
public static void main(String[] args){
int[] arr = new int[4];
recur(arr, 0);
}
}