// Source : https://leetcode.com/problems/nth-digit/ // Author : Hao Chen // Date : 2016-11-05 /*************************************************************************************** * * 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 31). * * 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. ***************************************************************************************/ #include using namespace std; class Solution { public: int findNthDigit(int n) { // We can see the following pattern: // // 1, 2, .... 9 : there are 9 * 1 digits. // 10, 11, ..., 99: there are 90 * 2 digits. // 101, 102, 103, ..., 999: there are 900 * 3. // ... //we can count the digits with the above pattern long digits_cnt = 0; long digits_cnt_prev = 0; int base = 0; for ( ; digits_cnt < n; base++) { digits_cnt_prev = digits_cnt; digits_cnt = digits_cnt + 9 * pow(10 , base) * ( base + 1 ); } // Now, we got `digits_cnt_prev`, `digits_cnt` and `base` // // For examples: // n = 20; digits_cnt_prev = 9, digits_cnt = 9+90*2 = 189, base = 2; // n = 500; digits_cnt_prev = 9+90*2 = 189, digits_cnt = 9+90*2+900*3 = 2889, base = 3; // n = 2000; digits_cnt_prev = 9+90*2 = 189, digits_cnt = 9+90*2+900*3 = 2889, base = 3; // // It means, we found the range where the number it is // n = 20, the number located in the range 10 -- 99 // n = 500, the number located in the range 100 - 999 // // and we can use `digits_cnt_prev` to know the previous rangs produce how many digits. // n = 20, the previous ranges produce 9 digits, so there needs 20-9 = 11 digits in [10 - 99] // n = 500, the previous ranges produce 189 digits, so there needs 500-189 = 311 digits in [100-999] // // the `base` told us in current ranges, each number can have how many digits. // then we can locate the target number. // n = 20, // (n - digits_cnt_prev) / base = (20 - 9 ) / 2 = 5, so, [10 - 14] produces 10 digits (ZERO-based), // now, we have 1 digits left, it is the first digit of the target number 15. // // n = 500, // (n - digits_cnt_prev) / base = (500 - 189) / 3 = 103, so, [100 - 202] produces 309 digits(ZERO-based). // now, we have (500 - 189 - 309) = 2 digits left, it is the second digit of the target number 203. // // We can write the code now... // int target = pow(10, base-1) + (n - digits_cnt_prev) / base - 1; int left = n - digits_cnt_prev - (n - digits_cnt_prev) / base * base; //cout << "target = " << target << ", left = " << left << endl; //no digits left if ( left == 0 ) return (target) % 10; //still have some digits left, it should be in next number. target++; return int( target / pow(10, base - left) ) % 10; } };