357 Count Numbers with Unique Digits

357. Count Numbers with Unique Digits​
1. Question
Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.
Example:
Given n = 2, return 91. (The answer should be the total numbers in the range of 0 ≤ x < 100, excluding[11,22,33,44,55,66,77,88,99])
2. Implementation
(1) Backtracking
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class Solution {
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    public int countNumbersWithUniqueDigits(int n) {
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        boolean[] used = new boolean[10];
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        int[] count = new int[1];
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        getNumbersWithUniqueDigits(0, n, used, count);
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        // 加1因为要算上刚开始的0 
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        return 1 + count[0];
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    }
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​
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    public void getNumbersWithUniqueDigits(int curLevel, int depth, boolean[] used, int[] count) {
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        if (curLevel == depth) {
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            return;
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        }
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​
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        int start = curLevel == 0 ? 1 : 0;
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​
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        for (int i = start; i < 10; i++) {
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            if (!used[i]) {
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                used[i] = true;
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                ++count[0];
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                getNumbersWithUniqueDigits(curLevel + 1, depth, used, count);
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                used[i] = false;
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            }
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        }
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    }
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}
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(2) DP
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class Solution {
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    public int countNumbersWithUniqueDigits(int n) {
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        if (n > 10) {
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            return 0;
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        }
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​
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        int[] dp = new int[11];
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        // 0个digit的时候，unique number是1， 这里我们把0当做
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        dp[0] = 1;
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        dp[1] = 9;
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​
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        for (int i = 2; i <= n; i++) {
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            dp[i] = dp[i - 1] * (9 - i + 2);
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        }
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​
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        int res = 0;
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        for (int i = 0; i <= n; i++) {
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            res += dp[i];
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        }
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        return res;
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    }
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}
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3. Time & Space Complexity
Backtracking: 时间复杂度O(n!), 如果n > 10的话，时间复杂度为O(10!), 空间复杂度O(n)
DP: 时间复杂度O(n), 空间复杂度O(1)

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Contents

357. Count Numbers with Unique Digits

1. Question

2. Implementation

3. Time & Space Complexity

357. Count Numbers with Unique Digits​

1. Question

2. Implementation

3. Time & Space Complexity

357. Count Numbers with Unique Digits