357. Count Numbers with Unique Digits # 题目 # Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n. Example: Input: 2 Output: 91 Explanation: The answer should be the total numbers in the range of 0 ≤ x < 100, e
leetcode| Count Numbers with Unique Digits 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...
leetcode-357-Count Numbers with Unique Digits 此题的总结: 求解 最大爆破值, 是一个 倒序 二分法问题,最终的原子结构是连续的三个数。 连续的三个数,可以 往上递推 间隔一个数的三个数,间隔n个数的三个数 特点在于:每一次递推,都有可能改变当前槽位值,因为,i,j不变,由于间隔变化,变得是取得间隔点。
leetcode [357]Count Numbers with Unique Digits Given anon-negativeinteger n, count all numbers with unique digits, x, where 0 ≤ x < 10n. Example: Input:2Output:91Explanation:The answer should be the total numbers in the range of 0 ≤ x < 100, excluding11,22,33,44,55,66,77,88,99...
LeetCode Count Numbers with Unique Digits 原题链接在这里:https://leetcode.com/problems/count-numbers-with-unique-digits/description/ 题目: Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.
leetCode 357. Count Numbers with Unique Digits | Dynamic Programming | Medium,357.CountNumberswithUniqueDigitsGivena non-negative integern,countallnumberswithuniquedigits,x,where0≤x<10n.Example:Givenn=2,return91.(Theanswershouldbethet
classSolution {public:intcountNumbersWithUniqueDigits(intn) {if(n ==0)return1;intres =0;for(inti =1; i <= n; ++i) {res+=count(i); }returnres; }intcount(intk) {if(k <1)return0;if(k ==1)return10;intres =1;for(inti =9; i >= (11- k); --i) { ...
int countNumbersWithUniqueDigits(int n) { vector<int> dp(11); dp[0] = 1; for(int i = 1; i <= 10; i ++) { int temp = 9; for(int j = 9; j > 10 - i; j--) { temp *= j; } dp[i] = temp + dp[i - 1]; } if(n > 10) return dp[10]; else return dp[n]...
357 Count Numbers with Unique Digits 计算各个位数不同的数字个数,给定一个非负整数n,计算各位数字都不同的数字x的个数,其中0≤x<10n。示例:给定n=2,返回91。(答案应该是除[11,22,33,44,55,66,77,88,99]外,0≤x<100间的所有数字)详见:https://leetcode.com/proble
357. Count Numbers with Unique Digits 难度:m n正的时候考虑[10^(n-1),10^n)之间的个数。如果没有0就是9*8...*(10-n),有0就是9*8*...*(11-n)*(n-1),和为9*8*...*(11-n)*9。所以只要挨个计算再相加就行了。 class Solution: ...