Enter the number n, the maximum print from 1 to n-bit decimal order. For example, input 3, then print out up to a maximum of 1, 2, 3 digits 999.
Example 1:
Input: n = 1
Output: [1,2,3,4,5,6,7,8,9]
Description:
by returning a list of integers instead of printing
n is a positive integer
In large numbers for the solution of
the lc written by one person. . . True a little difficult to understand
class Solution {
public:
vector<int> res;
vector<int> printNumbers(int n) {
if (n <= 0) return res;
//创建一个能容纳最大值的字符数组,由于有一位要存储'\0'，因此要开大一格
char* number = new char[n + 1];
//初始全部设置为0
memset(number, '0', n);
number[n] = '\0';//第n位设为'\0'
while (!Increment(number)){
PrintNumber(number);
}
delete[]number;//注意要释放内存
return res;
}
bool Increment(char* number) {//形参传递char*指针
bool isOverflow = false;//检测是否越界
int nTakeOver = 0;//存储进位
int nLength = strlen(number);//长度为n,不是n+1
for (int i = nLength - 1; i >= 0; i--){
int nSum = number[i] - '0' + nTakeOver;
if (i == nLength - 1){
//如果是第一位，进位
nSum++;
}
if (nSum >= 10){
//有进位
if (i == 0){
//如果是最高位有进位，说明超过了给定得到最大值，越界
isOverflow = true;
}
else{
//非最高位有进位
nTakeOver = 1;
number[i] = nSum - 10 + '0';//对第i位进行设置
}
}
else{
//没有进位
//设置第i位数字
//并直接跳出循环
number[i] = nSum + '0';
break;
}
}
return isOverflow;
}
void PrintNumber(char* number){
//形参传递char*指针，此处改变形参number指向的位置，不会使原始的number指针所指位置改变
string s = "";
bool isBegin0 = true;
while (*number != '\0'){
if (isBegin0 && *number != '0') isBegin0 = false;
//碰到'0'，则不输出
if (!isBegin0){
s += *number;
}
number++;
}
int num = stoi(s);//转为整数
res.push_back(num);
}
};