Lintcode: Fast Power

Calculate the a^n % b where a, b and n are all 32bit integers.

Example
For 2^31 % 3 = 2

For 100^1000 % 1000 = 0

Challenge
O(logn)

这道题跟Pow这道题很像

数学问题，要求O(log n)的时间复杂度，也就是每次去掉一半的计算量，先要找到对应的数学公式：
(a * b) % p = (a % p * b % p) % p

所以(a^n)%b = (a^n/2 * a^n/2 * a) % b = ((a^n/2 * a^n/2)%b * a%b) % b
注意int和long的转化，防止溢出。

比较严谨的做法：

class Solution {
    /*
     * @param a, b, n: 32bit integers
     * @return: An integer
     */
    public int fastPower(int a, int b, int n) {
        // write your code here
        long ret = getPower(a, b, n);
        return (int)ret;
    }
    public long getPower(int a, int b, int n){
        if(a == 0) return 0;
        if(n == 0) return 1 % b;
        if(n == 1) return a % b;
         
        long ret = getPower(a, b, n/2);
        ret *= ret;
        ret %= b;
        if(n % 2 == 1){
            ret = ret * (a % b);
        }
        return ret % b;
    }
};

我当时的做法：

class Solution {
    public int fastPower(int a, int b, int n) {
        if (n == 0) return 1%b;
        long half = fastPower(a, b, n/2);
        if (n % 2 == 0) {
            long temp = half * half;
            return (int)(temp % b);
        }
        else {
            long temp = (half * half)%b * (a%b);
            return (int)(temp % b);
        }
    }
};

class Solution {
    /*
     * @param a, b, n: 32bit integers
     * @return: An integer
     */
    public int fastPower(int a, int b, int n) {
        long ret = helper(a, b, n);
        return (int)ret;
    }
    
    public long helper(int a, int b, int n) {
        if (n == 0) return 1%b;
        long half = helper(a, b, n/2);
        if (n % 2 == 0) {
            return half * half % b;
        }
        else {
            return ((half * half % b) * a%b)%b;
        }
    }
};