[LintCode] Print Numbers by Recursion

kumfo 发布于2019-08-14 15:17 / 2762人阅读

摘要：只有当位数时，才打印数字。首先分析边界，应该，然后用存最高位。用函数对进行递归运算，同时更新结果数组。更新的过程归纳一下，首先，计算最高位存入，然后，用到倍的和之前里已经存入的所有的数个循环相加，再存入，更新，计算更高位直到等于

Problem

Print numbers from 1 to the largest number with N digits by recursion.

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].

Note

只有当位数n >= 0时，才打印数字。首先分析边界n = 0，应该return 1，然后用base存最高位。用helper()函数对base进行递归运算，同时更新结果数组res。
更新res的过程：

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...

Solution

Recursion

public class Solution {
    public List numbersByRecursion(int n) {
        List res = new ArrayList();
        if (n >= 0) {
            helper(n, res);
        }
        return res;
    }
    public int helper(int n, List 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;
    }
}

Non-recursion

public class Solution {
    public List numbersByRecursion(int n) {
        List res = new ArrayList();
        int max = 1;
        while (n != 0) {
            max *= 10;
            n--;
        }
        for (int i = 1; i < max; i++) {
            res.add(i);
        }
        return res;
    }
}

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