## Saturday, October 8, 2016

### LeetCode OJ - Nth Digit

Problem:

Analysis:

Note that within a block, the pattern is pretty regular. All we need is to make sure we know which block a digit is in. This can be done by checking them one by one

Solution:

The first block goes from 1 - 9, these are 9 1 digit numbers. Therefore there are 9 digits.
The second block goes from 10 - 99, these are 99 - 10 + 1 = 90 2 digit numbers. Therefore there are 180 digits.
The third block goes from 100 - 999, these are 999 - 100 + 1 = 900 3 digit numbers. Therefore there are 2700 digits.

The pattern should be fairly obviously now. Just check if the input is within the certain block, find the number out and do digit extraction.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/nth-digit/

#include "LEET_NTH_DIGIT.h"
#include <map>
#include <iostream>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_NTH_DIGIT
{
class Solution
{
public:
int findNthDigit(int n)
{
// Programmer reasons with 0 based index
n--;
long long blockBase = 1;
int blockValueSize = 1;
while (true)
{
// An exclusive index for the 'value' that is in the block
long long blockCeil = blockBase * 10;

// Number of digits in the block
long long blockSizeLong = (blockCeil - blockBase) * blockValueSize;

int intMax = ~(1 << 31);

int blockSize = blockSizeLong > intMax ? intMax : (int)blockSizeLong;

if (n < blockSize)
{
// We are now within a block

// What is the index in the block are we talking about?
int index = n / blockValueSize;

// Now we get the value of that number
long long value = blockBase + index;

// Which digit (are we talking about), this is 0 based, counting from most significant digit
int digit = n % blockValueSize;

// Change it to 1 based
digit++;

// Counting from the least significant digit instead
digit = blockValueSize - digit + 1;

// Digit extraction
while (digit > 1)
{
value /= 10;
digit--;
}
return value % 10;
}
else
{
// Move to the next block
n -= blockSize;
blockBase *= 10;
blockValueSize++;
}
}
}
};
};

using namespace _LEET_NTH_DIGIT;

int LEET_NTH_DIGIT()
{
Solution solution;
for (int i = 1; i <= 200; i++)
{
cout << solution.findNthDigit(i);
}
cout << endl;
return 0;
}