題目:
Find the nth digit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...
Note:
n is positive and will fit within the range of a 32-bit signed integer (n < 231).
Example 1:
Input:
3
Output:
3
Example 2:
Input:
11
Output:
0
Explanation:
The 11th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10.
解答:
//以10的n次方作為分界線,計算之前所有數字的總個數,與n值比較,确定所求結果所在範圍,然後再确定所求結果所在數值,得出結果
int findNthDigit(int n) {
int flag=1;
int multi=1;
long sum=9;
int mod;
long num;
if(n<10) return n;
while(sum<n)
{
flag++;
multi=multi*10;
if(flag==9)break;//sum 為int類型,防止sum超出存儲範圍
sum=sum+flag*(multi*10-multi);
}
if(flag==9)
{
mod=(n-sum)%flag;
}
else
{
sum=(sum-flag*(multi*10-multi));
mod=(n-sum)%flag;
}
if(mod==0)
{
num=(n-sum)/flag+multi-1;
num=num%10;
}
else
{
num=(n-sum)/flag+multi;
num=num/pow(10,flag-mod);
num=num%10;
}
return num;
}