357.[LeetCode]Count Numbers with Unique Digits
2016-08-31 00:32
417 查看
这个题我的思路有问题: 我一直想的都是计算出重复数然后用总数减去，简直是舍近求远，可以结合概率统计的原理直接求出无重复数的呀
public class Solution {
int countNumbersWithUniqueDigits(int n) {
if(n==0) return 1;
if(n==1) return 10;
int val = 9, ans = 10;
for(int i = 2; i <= n; i++)
{
val *= (9-i+2);
ans += val;
}
return ans;
}
}
内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理 
相关文章推荐
- 357.leetcode Count Numbers with Unique Digits(easy)[数学问题 非重复数字]
- [LeetCode] Count Numbers with Unique Digits
- LeetCode | Count Numbers with Unique Digits
- [LeetCode] Count Numbers with Unique Digits 计算各位数值不同的数的个数
- leetcode 357 c++. Count Numbers with Unique Digits
- LeetCode Count Numbers with Unique Digits(计数问题)
- leetcode:数学：Count Numbers with Unique Digits(357)
- LeetCode 357 Count Numbers with Unique Digits
- Count Numbers with Unique Digits -- LeetCode
- [LeetCode]Count Numbers with Unique Digits
- 【LEETCODE】357- Count Numbers with Unique Digits [Python]
- Leetcode 357 Count Numbers with Unique Digits
- LeetCode--No.357--Count Numbers with Unique Digits
- LeetCode 357 Count Numbers with Unique Digits (排列组合)
- leetcode Count Numbers with Unique Digits
- Leetcode: Count Numbers with Unique Digits
- leetcode：Count Numbers with Unique Digits
- LeetCode No.357 Count Numbers with Unique Digits
- LeetCode: Count Numbers with Unique Digits
- LeetCode: Count Numbers with Unique Digits