資訊專欄INFORMATION COLUMN
摘要：只有當(dāng)位數(shù)時(shí)，才打印數(shù)字。首先分析邊界，應(yīng)該，然后用存最高位。用函數(shù)對(duì)進(jìn)行遞歸運(yùn)算，同時(shí)更新結(jié)果數(shù)組。更新的過程歸納一下，首先，計(jì)算最高位存入，然后，用到倍的和之前里已經(jīng)存入的所有的數(shù)個(gè)循環(huán)相加，再存入，更新，計(jì)算更高位直到等于
Problem
Print numbers from 1 to the largest number with N digits by recursion.
ExampleGiven 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只有當(dāng)位數(shù)n >= 0時(shí)，才打印數(shù)字。首先分析邊界n = 0，應(yīng)該return 1，然后用base存最高位。用helper()函數(shù)對(duì)base進(jìn)行遞歸運(yùn)算，同時(shí)更新結(jié)果數(shù)組res。
更新res的過程：
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;
//...
歸納一下，首先，計(jì)算最高位存入base，然后，用1到9倍的base（curbase）和之前res里已經(jīng)存入的所有的數(shù)（res.size()個(gè)）循環(huán)相加，再存入res，更新res.size，計(jì)算更高位直到base等于10^n...
SolutionRecursion
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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