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