New Problem Solution "Nth Digit"

Author: haoel

README.md

@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|400|[Nth Digit](https://leetcode.com/problems/nth-digit/) | [C++](./algorithms/cpp/nthDigit/NthDigit.cpp)|Easy|
 |399|[Evaluate Division](https://leetcode.com/problems/evaluate-division/) | [C++](./algorithms/cpp/evaluateDivision/EvaluateDivision.cpp)|Medium|
 |398|[Random Pick Index](https://leetcode.com/problems/random-pick-index/) | [C++](./algorithms/cpp/randomPickIndex/RandomPickIndex.cpp)|Medium|
 |397|[Integer Replacement](https://leetcode.com/problems/integer-replacement/) | [C++](./algorithms/cpp/integerReplacement/IntegerReplacement.cpp)|Medium|

algorithms/cpp/nthDigit/NthDigit.cpp

@@ -0,0 +1,99 @@
+// 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 <cmath>
+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;
+    }
+};