LintCode 371 Print Numbers by Recursion


https://www.lintcode.com/en/problem/print-numbers-by-recursion/
Print numbers from 1 to the largest number with N digits by recursion.
 Notice
It's pretty easy to do recursion like:
recursion(i) {
    if i > largest number:
        return
    results.add(i)
    recursion(i + 1)
}
however this cost a lot of recursion memory as the recursion depth maybe very large. Can you do it in another way to recursive with at most N depth?
Example
Given N = 1, return [1,2,3,4,5,6,7,8,9].
Given N = 2, return [1,2,3,4,5,6,7,8,9,10,11,12,...,99].
从小至大打印 N 位的数列,正如题目中所提供的 recursion(i), 解法简单粗暴,但问题在于 N 稍微大一点时栈就溢出了,因为递归深度太深了。能联想到的方法大概有两种,一种是用排列组合的思想去解释,把0~9当成十个不同的数(字符串表示),塞到 N 个坑位中,这个用 DFS 来解应该是可行的;另一个则是使用数学方法,依次递归递推,比如 N=2 可由 N=1递归而来,具体方法则是乘10进位加法。题中明确要求递归深度最大不超过 N, 故 DFS 方法比较危险。
https://zhengyang2015.gitbooks.io/lintcode/print_numbers_by_recursion_371.html
递归求解n时,可以对n-1位置上的数从0-9遍历,依次类推直到最后一位。
    public List<Integer> numbersByRecursion(int n) {
        List<Integer> res = new ArrayList<Integer>();
        if(n < 1){
            return res;
        }

        helper(n, 0, res);
        return res;
    }

    //ans在这里可以看成前一位的值,在递归过程中会乘以n-1次10,即10^(n-1)
    private void helper(int n, int ans, List<Integer> res){
        if(n==0){
            if(ans>0){
                res.add(ans);
            }
            return;
        }

        int i;
        for(i=0; i<=9; i++){
            helper(n-1, ans*10+i, res);
        }
    }
res = {1, 2, 3, 4, 5, 6, 7, 8, 9};
base = helper(n-1, res); //10
//i = 1 ~ 9;
for i:
curbase = i * base; //10, 20, 30, 40, ... , 80, 90
res.add(curbase); // 10; 20; ... ; 80; 90
//j = index of res;
for j:
res.add(curbase + res.get(j)); //11, 12, 13, ... , 18, 19;
                               //21, 22, 23, ... , 28, 29;
                               //...
归纳一下,首先,计算最高位存入base,然后,用1到9倍的base(curbase)和之前res里已经存入的所有的数(res.size()个)循环相加,再存入res,更新res.size,计算更高位直到base等于10^n...
    public List<Integer> numbersByRecursion(int n) {
        List<Integer> res = new ArrayList<Integer>();
        if (n >= 0) {
            helper(n, res);
        }
        return res;
    }
    public int helper(int n, List<Integer> res) {
        if (n == 0) return 1;
        int base = helper(n-1, res);
        int size = res.size();
        for (int i = 1; i <= 9; i++) {
            int curbase = i * base;
            res.add(curbase);
            for (int j = 0; j < size; j++) {
                res.add(curbase + res.get(j));
            }
        }
        return base * 10;
    }
递归步的截止条件n == 0, 由于需要根据之前 N-1 位的数字递推,base 每次递归一层都需要乘10,size需要在for循环之前就确定。
+
    public List<Integer> numbersByRecursion(int n) {
        List<Integer> result = new ArrayList<Integer>();
        if (n <= 0) {
            return result;
        }
        helper(n, result);
        return result;
    }

    private void helper(int n, List<Integer> ret) {
        if (n == 0) return;
        helper(n - 1, ret);
        // current base such as 10, 20, 30...
        int base = (int)Math.pow(10, n - 1);
        // get List size before for loop
        int size = ret.size();
        for (int i = 1; i < 10; i++) {
            // add 10, 100, 1000...
            ret.add(i * base);
            for (int j = 0; j < size; j++) {
                // add 11, 12, 13...
                ret.add(ret.get(j) + base * i);
            }
        }
    }
n与n-1的关系其实就是将n-1的所有的结果,加上10 * (n - 1) i (i from 1 - 9) 但是对于2和1不一样,因为1的结果不应该包括0, 所以要单独考虑
+

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