你可不可以
成为我的main函数
做我此生必须有
且只能有一个的入口
——————————
我愿为自己加上private
在你的class中只有
你能调用
——————————————代 码 如 诗 。
Given an integer, write a function to determine if it is a power of three.
Follow up:
Could you do it without using any loop / recursion?
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class Solution { public: bool isPowerOfThree(int n) { if(n <= 0) return false; return pow(3, (round)(log(n) / log(3))) == n; } }; |
Given an integer, write a function to determine if it is a power of two.
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update v2.0: class Solution { public: bool isPowerOfTwo(int n) { if(n <= 0) return false; return pow(2, (round)(log(n) / log(2))) == n; } }; //如果没有考虑n <= 0就会超时。。 class Solution { public: bool isPowerOfTwo(int n) { if(n <= 0) return false; while(n != 1) { if(n % 2 == 0) { n = n / 2; } else { return false; } } return true; } }; |
Given an array and a value, remove all instances of that value in place and return the new length.
Do not allocate extra space for another array, you must do this in place with constant memory.
The order of elements can be changed. It does not matter what you leave beyond the new length.
Example:
Given input array nums = [3,2,2,3], val = 3
Your function should return length = 2, with the first two elements of nums being 2.
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class Solution { public: int removeElement(vector<int>& nums, int val) { int len = 0; for(int i = 0; i < nums.size(); i++) { if(nums[i] != val) { nums[len++] = nums[i]; } } return len; } }; |
Write an algorithm to determine if a number is “happy”.
A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.
Example: 19 is a happy number
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1
Credits:
Special thanks to @mithmatt and @ts for adding this problem and creating all test cases.
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//即使输入的数为9*10^10，也会在第一次 while 降到3位数，第二次 while 降到2位数，不超过10次肯定能知道是否满足题目条件 //所以加了一个 cnt 计数器 在超过10次就跳出循环说明不满足题意 class Solution { public: bool isHappy(int n) { int cnt = 0; while(n != 1) { cnt++; int temp = 0; while(n) { temp = temp + (n % 10) * (n % 10); n = n / 10; } n = temp; if(cnt == 10) { return false; } } return true; } }; |
Write a program to check whether a given number is an ugly number.
Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 6, 8 are ugly while 14 is not ugly since it includes another prime factor 7.
Note that 1 is typically treated as an ugly number.
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class Solution { public: bool isUgly(int num) { if(num <= 0) { return false; } int flag = 0; while(num != 1) { flag = 0; if(num % 2 == 0) { num = num / 2; flag = 1; } if(num % 3 == 0) { num = num / 3; flag = 1; } if(num % 5 == 0) { num = num / 5; flag = 1; } if(flag == 0) { return false; } } return true; } }; |
You are climbing a stair case. It takes n steps to reach to the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
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class Solution { public: int climbStairs(int n) { int *a = new int [n + 1]; a[0] = 1; a[1] = 1; for(int i = 2; i <= n; i++) { a[i] = a[i - 1] + a[i - 2]; } return a[n]; } }; //一开始我写了个Fibonacci递归，在n = 44 时候超时了 class Solution { public: int climbStairs(int n) { if(n == 1 || n == 2) return n; return climbStairs(n - 1) + climbStairs(n - 2); } }; //最最一开始我用的最原始的递归， 在 n=38的时候超时了 class Solution { public: int ans = 0; int climbStairs(int n) { if(n == 0) { ans++; return ans; } if(n >= 1) { climbStairs(n - 1); } if(n >= 2) { climbStairs(n - 2); } } }; |