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FastPower.java
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91 lines (69 loc) · 1.88 KB
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package Algorithms.lintcode.math;
class FastPower {
public static void main(String[] strs) {
System.out.println(fastPower(2147483647, 2147483645, 214748364));
//System.out.println(Math.pow(2, 2147483647));
System.out.println(9 % (-2));
}
/*
* @param a, b, n: 32bit integers
* @return: An integer
*/
/*
* @param a, b, n: 32bit integers
* @return: An integer
*/
public static int fastPower(int a, int b, int n) {
// write your code here
long ret = pow(a, b, n);
return (int) ret;
}
// suppose n > 0
public static long pow1(int a, int b, int n) {
if (a == 0) {
return 0;
}
// The base case.
if (n == 0) {
return 1 % b;
}
if (n == 1) {
return a % b;
}
long ret = 0;
// (a * b) % p = (a % p * b % p) % p （3）
ret = pow(a, b, n / 2);
ret *= ret;
// 这一步是为了防止溢出
ret %= b;
if (n % 2 == 1) {
ret *= pow(a, b, 1);
}
// 执行取余操作
ret = ret % b;
return ret;
}
// SOLUTION 2:
// suppose n > 0
public static long pow(int a, int b, int n) {
if (a == 0) {
return 0;
}
// The base case.
if (n == 0) {
return 1 % b;
}
long ret = 0;
// (a * b) % p = (a % p * b % p) % p （3）
ret = pow(a, b, n / 2);
ret *= ret;
// 这一步是为了防止溢出
ret %= b;
if (n % 2 == 1) {
ret *= (a % b);
}
// 执行取余操作
ret = ret % b;
return ret;
}
};