一、题目描述

输入数字 n，按顺序打印出从 1 到最大的 n 位十进制数。比如输入 3，则打印出 1、2、3 一直到最大的 3 位数 999。
示例 1:
输入: n = 1
输出: [1,2,3,4,5,6,7,8,9]
说明：
用返回一个整数列表来代替打印
n 为正整数

二、题解
全排列

class Solution {
    
    
    private int[] res;
    private char[] loop = {
    
    '0','1','2','3','4','5','6','7','8','9'};
    private char[] num;
    private int n;
    private int count = 0;
    public int[] printNumbers(int n) {
    
    
        this.n = n;
        this.num = new char[n];
        this.res = new int[(int)Math.pow(10,n)-1];

        dfs(0);

        return res;
    }
    public void dfs(int x){
    
    
        
        if(x == n){
    
    
            if(Integer.parseInt(String.valueOf(num))!=0)
                res[count++] = Integer.parseInt(String.valueOf(num));
            return ;
        }
        for(char ch:loop){
    
    
            num[x] = ch;
            dfs(x + 1);
        }
    }
}

删除高位多余的 00,(由于上题是转化为了整数，如果是字符串的话，会出现001这样的情况，因此需要解决)

class Solution {
    
    
    private int[] res;
    private char[] loop = {
    
    '0','1','2','3','4','5','6','7','8','9'};
    private char[] num;
    private int n;
    private int count = 0;
    private int nine = 0;
    private int start = 0;
    public int[] printNumbers(int n) {
    
    
        this.n = n;
        this.num = new char[n];
        this.res = new int[(int)Math.pow(10,n)-1];
        this.start = n - 1;

        dfs(0);

        return res;
    }
    public void dfs(int x){
    
    
        
        if(x == n){
    
    
            int temp = Integer.parseInt(String.valueOf(num).substring(start));
            if(temp!=0)
                res[count++] = temp;
            if(n-start == nine) start--;
            return ;
        }
        for(char ch:loop){
    
    
            if(ch == '9') nine++;
            num[x] = ch;
            dfs(x + 1);
        }
        nine--;
    }
}