New Problem Solution "Count Numbers with Unique Digits"

README.md

@@ -107,6 +107,7 @@ LeetCode
 |376|[Wiggle Subsequence](https://leetcode.com/problems/wiggle-subsequence/) | [C++](./algorithms/cpp/wiggleSubsequence/wiggleSubsequence.cpp)|Medium|
 |371|[Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/description/) | [C++](./algorithms/cpp/sumOfTwoIntegers/SumOfTwoIntegers.cpp)|Easy|
 |367|[Valid Perfect Square](https://leetcode.com/problems/valid-perfect-square/description/) | [C++](./algorithms/cpp/validPerfectSquare/ValidPerfectSquare.cpp)|Easy|
+|357|[Count Numbers with Unique Digits](https://leetcode.com/problems/count-numbers-with-unique-digits/) | [C++](./algorithms/cpp/countNumbersWithUniqueDigits/CountNumbersWithUniqueDigits.cpp)|Medium|
 |350|[Intersection of Two Arrays II](https://leetcode.com/problems/intersection-of-two-arrays-ii/) | [C++](./algorithms/cpp/intersectionOfTwoArrays/intersectionOfTwoArraysII.cpp)|Easy|
 |349|[Intersection of Two Arrays](https://leetcode.com/problems/intersection-of-two-arrays/) | [C++](./algorithms/cpp/intersectionOfTwoArrays/intersectionOfTwoArrays.cpp)|Easy|
 |347|[Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements/) | [C++](./algorithms/cpp/topKFrequentElements/topKFrequentElements.cpp)|Medium|

algorithms/cpp/countNumbersWithUniqueDigits/CountNumbersWithUniqueDigits.cpp

@@ -0,0 +1,70 @@
+// Source : https://leetcode.com/problems/count-numbers-with-unique-digits/
+// Author : Hao Chen
+// Date   : 2019-03-24
+
+/***************************************************************************************************** 
+ *
+ * Given a non-negative integer n, count all numbers with unique digits, x, where 0 &le; x < 10n.
+ * 
+ * Example:
+ * 
+ * Input: 2
+ * Output: 91 
+ * Explanation: The answer should be the total numbers in the range of 0 &le; x < 100, 
+ *              excluding 11,22,33,44,55,66,77,88,99
+ * 
+ ******************************************************************************************************/
+
+
+// Considering three digits
+//   - the first place could be [1-9] which has 9 choices.
+//   - the second place could be [0-9] with excluding the first digit, which is 10-1=9 choices.
+//   - the third place could be [0-9] with excluding the 1st and 2nd digits, which is 10-2=8 choices.
+// So, three digits has 9*9*8 unique digits.
+//
+// After adds the 1 digit unique number,and 2 digits unique number, we can have the result:
+//
+//   9*9*8 + 9*9 + 10 = 648 + 81 + 10 = 739
+//
+// n = 0, a[0] = 1;
+// n = 1, a[1] = 9 + a[0];
+// n = 2, a[2] = 9*9 + a[1];
+// n = 3, a[3] = 9*9*8 + a[2];
+// n = 4, a[4] = 9*9*8*7 + a[3];
+// ....
+class Solution {
+public:
+    int countNumbersWithUniqueDigits(int n) {
+        int result = 1;
+        for (int i=0; i<n; i++) {
+            result += ( 9 * nine_factor(i) );
+        }
+        return result;
+    }
+
+    int nine_factor(int n) {
+        int result = 1;
+        for (int i=0; i<n; i++) {
+            result *= (9-i);
+        }
+        return result;
+    }
+};
+
+
+//actually, the function nine_factor() could be optimized!
+
+class Solution {
+public:
+    int countNumbersWithUniqueDigits(int n) {
+        int result = 1;
+        int nine_factor = 1;
+        for (int i=0; i<n; i++) {
+            result += ( 9 * nine_factor );
+            nine_factor *= (9-i);
+        }
+        return result;
+    }
+
+
+};