- Leetcode 357.Count Number With U
- 357. Count Numbers with Unique D
- 357. Count Numbers with Unique D
- 357. Count Numbers with Unique D
- 357. Count Numbers with Unique D
- 357. Count Numbers with Unique D
- 357. Count Numbers with Unique D
- 357. Count Numbers with Unique D
- 357. Count Numbers with Unique D
- Leetcode 357. Count Numbers with
解题思路：
- n的数量= n-1的数量+9*8****。。。。。
class Solution {
public int countNumbersWithUniqueDigits(int n) {
if(n == 0) return 1;
else if(n == 1) return 10;
int[] dp = new int[n + 1];
dp[0] = 1;
dp[1] = 10;
for(int i = 2; i <= n; i++){
dp[i] = 9;
int k = 9;
int count = 1;
while(count < i){
dp[i] *= k;
count++;
k--;
}
dp[i] += dp[i - 1];
}
return dp[n];
}
}
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