leetcode简单题目两道(5)

Problem

Given an integer (signed 32 bits), write a function to check whether it is a power of 4.

Example:

Given num = 16, return true. Given num = 5, return false.

Follow up: Could you solve it without loops/recursion?

Code

class Solution {
public:
bool isPowerOfFour(int num) {
return (num > 0) && ((num & (num - 1)) == 0)  && ((num - 1) % 3 == 0);
}
};
说明

利用了2的指数与本身减1相与为0，以及4的指数减1，必定能整除3；

class Solution {
public:
bool isPowerOfFour(int num) {
return (num > 0) && ((num & (num - 1)) == 0)  && ((num & 0x55555555));
}
};
说明

利用了2的指数与本身减1相与为0，以及4的指数在16进制中的位置0x55555555，前者确定只有一个1，后者确定这个1肯定是4的指数的位置；

problem

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.

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.

Code

class Solution {
public:
string int2str(int int_temp)
{
stringstream stream;
stream << int_temp;
return stream.str();
}
string getHint(string secret, string guess) {
if(secret.size() == 0 || guess.size() == 0) {
return 0;
}
int res1 = 0, res2 = 0, tmp;
map<char, int> map1, map2;
for(int i = 0; i < secret.size(); i++) {
if (secret[i] == guess[i]) {
res1++;
} else {
map1[secret[i]]++;
map2[guess[i]]++;
}
}
map<char,int>::iterator it;
for(it=map1.begin();it!=map1.end();++it)
{
if (it->second < map2[it->first]) {
tmp = it->second;
} else {
tmp = map2[it->first];
}
res2 += tmp;
}
return int2str(res1) + "A" + int2str(res2) + "B";
}
};
一次遍历，不解释，哈哈。