Given a positive integer, return its corresponding column title as appear in an Excel sheet.
For example:

    1 -> A
    2 -> B
    3 -> C
    ...
    26 -> Z
    27 -> AA
    28 -> AB

这题看起来不难，但居然做了我两个小时。其中还是有很多值得注意一下的地方的。

def convertToTitle2(n):
    ret = ""
    while(n):
        ret = chr( (n-1)%26 + 65 ) + ret
        n = (n-1)/26 #核心
    return ret

先贴一下最后AC的代码。我之前用了很长的代码去判断字符进位的情况，但效果都不理想。

Given an integer, write a function to determine if it is a power of three.

class Solution(object):
    def isPowerOfThree(self, n):
        #while (n and (n % 3 == 0)):
            #n = n / 3
        #return n == 1
        
        
        #return (n > 0 and 1162261467 % n == 0)
        
        return (n > 0 and int(math.log10(n) / math.log10(3)) - math.log10(n) / math.log10(3) == 0)

Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode* root)
    {
        if(root == NULL)
            return 0;  
        int res = 1;  
        int l = maxDepth(root->left);  
        int r = maxDepth(root->right);  
        return l > r? l + 1:r+ 1;  
    }
};

Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct.

class Solution {
public:
    bool containsDuplicate(vector<int>& nums)
    {
        if(nums.empty())
            return false;
            
        set<int> s;
        vector<int>::iterator itr = nums.begin();
        
        while(itr != nums.end())
        {
            if(s.count(*itr) == 1)
                return true;
            s.insert(*itr);
            itr++;
        }
        return false;
    }
};

Given an integer, write a function to determine if it is a power of two.

class Solution(object):
    def isPowerOfTwo(self, n):
        while (n and (n % 2 == 0)):
            n = n / 2
            
        return n == 1
        
        #Certainly,you can also use the following method
        #return (n and (n&(n-1))) == 0

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.

class Solution(object):
    def isUgly(self, number):
        if number == 1:
            return True
        if number == 0:
            return False
        
        while(number % 2 == 0):
            number /= 2
        while(number % 3 == 0):
            number /= 3
        while(number % 5 == 0):
            number /= 5
    
        return number == 1

Write a program to find the n-th ugly number.
Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.
Note that 1 is typically treated as an ugly number.

class Solution(object):
    def min_three(self, a, b, c):
        minNum = min(a,b)
        return min(minNum,c)
    def nthUglyNumber(self, n):
        ugly = [1]
        factor2 = 2
        factor3 = 3
        factor5 = 5
        index2 = index3 = index5 = 0
        for i in range(1,n):
            minNum = min(factor2, factor3, factor5)
            ugly.append(minNum)
            if(factor2 == minNum):
                factor2 = 2 * ugly[index2+1]
                index2 += 1
            if(factor3 == minNum):
                factor3 = 3 * ugly[index3+1]
                index3 += 1
            if(factor5 == minNum):
                factor5 = 5 * ugly[index5+1]
                index5 += 1
        return ugly[n-1]

You are playing the following Bulls and Cows game with your friend: You write down a number and ask your friend to guess what the number is. Each time your friend makes a guess, you provide a hint that indicates how many digits in said guess match your secret number exactly in both digit and position (called “bulls”) and how many digits match the secret number but locate in the wrong position (called “cows”). Your friend will use successive guesses and hints to eventually derive the secret number.
For example:

Secret number: “1807”
Friend’s guess: “7810”

Hint: 1 bull and 3 cows. (The bull is 8, the cows are 0, 1 and 7.)
Write a function to return a hint according to the secret number and friend’s guess, use A to indicate the bulls and B to indicate the cows. In the above example, your function should return “1A3B”.
Please note that both secret number and friend’s guess may contain duplicate digits, for example:

Secret number: “1123”
Friend’s guess: “0111”

In this case, the 1st 1 in friend’s guess is a bull, the 2nd or 3rd 1 is a cow, and your function should return “1A1B”.
You may assume that the secret number and your friend’s guess only contain digits, and their lengths are always equal.

class Solution {
public:
    string getHint(string secret, string guess)
    {
        int cntA = 0, cntB = 0;
        unordered_map<char, int> hash;
        vector<bool> tag(secret.size(), false);
        
        for (auto a : secret)
        {
            ++hash[a];
        };
        
        for (int i = 0; i < secret.size(); ++i)
        {
            if (secret[i] == guess[i])
            {
                ++cntA;
                --hash[secret[i]];
                tag[i] = true;
            }
        }
        for (int i = 0; i < guess.size(); ++i)
        {
            if (!tag[i] && hash[guess[i]] > 0)
            {
                ++cntB;
                --hash[guess[i]];
            }
        }
        return to_string(cntA) + "A" + to_string(cntB) + "B";
    }
};