Count the number of k's between 0 and n. k can be 0 - 9. Example if n = 12, k = 1 in [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] we have FIVE 1's (1, 10, 11, 12) 分析: 利用while 循环 + num % 10可以获取 num中的所有数字. class Solution { /* * param k : As description.…

Lesson learnt: one effective solution for bit\digit counting problems: counting by digit\bit http://www.hawstein.com/posts/20.4.html class Solution { public: /* * param k : As description. * param n : As description. * return: How many k's between 0…

Description Count the number of k's between 0 and n. k can be 0 - 9. Example if n = 12, k = 1 in [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] we have FIVE 1's (1, 10, 11, 12) Answer /** * @param k: An integer * @param n: An integer * @return: An intege…

题目链接 The 5-digit number, 16807=75, is also a fifth power. Similarly, the 9-digit number, 134217728=89, is a ninth power. How many n-digit positive integers exist which are also an nth power? 这个题目有点坑: 先说自己的思路<虽然方法不是很好> 根据题意可知道: a的b次方 除以 最小的b位数(如:1,10…

The 5-digit number, 16807=75, is also a fifth power. Similarly, the 9-digit number, 134217728=89, is a ninth power. How many n-digit positive integers exist which are also an nth power? 这种数字满足下面条件: 对于数位为x的数S=k^x 有 10^(x-1)<=k^x<=10^x-1 #include &quo…

Count the number of k's between 0 and n. k can be 0 - 9.   Example if n = 12, k = 1 in [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] we have FIVE 1's (1, 10, 11, 12) 解法一: class Solution { public: // param k : description of k // param n : description of…

题目: Count the number of k's between 0 and n. k can be 0 - 9. Example if n = 12, k = 1 in [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] we have FIVE 1's (1, 10, 11, 12) 题解: Solution 1 () class Solution { public: int digitCounts(int k, int n) { ) { ; } ;…

五位数\(16807=7^5\)也是一个五次幂,同样的,九位数\(134217728=8^9\)也是一个九次幂.求有多少个\(n\)位正整数同时也是\(n\)次幂? 分析:设题目要求的幂的底为\(n\),指数为\(k\),则这个幂应为\(k\)位数,则有: \[ 10^{k-1}<n^k<10^k \Rightarrow k-1<k\cdot log_{10}n<k \] 因为\(k\ge1\),则对于不等式\(k\cdot log_{10}n<k\),有\(log_{10}…

Yet Another Source Code for LintCode Current Status : 232AC / 289ALL in Language C++, Up to date (2016-02-10) For more problems and solutions, you can see my LintCode repository. I'll keep updating for full summary and better solutions. See cnblogs t…

面试题3 -- 搜索二维矩阵 写出一个高效的算法来搜索 m × n矩阵中的值. 这个矩阵具有以下特性: 1. 每行中的整数从左到右是排序的. 2. 每行的第一个数大于上一行的最后一个整数. public class Solution { /** * @param matrix, a list of lists of integers * @param target, an integer * @return a boolean, indicate whether matrix contains…