Prove safety offer: integer power (leetcode 50) value

topic:

Given a double base type and floating-point type int integer exponent. Seeking exponent of the power base.

Ensure the base and exponent are not simultaneously 0

answer:

Solution one:

Direct use of library functions Math.pow (), the following code:

public class Solution {
    public double Power(double base, int exponent) {
        return Math.pow(base,exponent);
  }
}

Solution two:

Using a loop, one by one, to pay attention to: determine whether the index is negative; Note exponent overflow; code as follows:

public class Solution {
    public double Power(double base, int exponent) {
        if(exponent==0){
            return 1;
        }
        boolean flag = (exponent>0);
        long exp = (long)Math.abs((long)exponent);
        double result = 1;
        for(long i=0;i<exp;i++){
            result = result*base;
        }
        if(!flag){
            result = 1/result;
        }
        return result;
  }
}

Solution three:

Dichotomy, as follows:

public class Solution {
    public double Power(double base, int exponent) {
        if(exponent==0){
            return 1;
        }
        boolean flag = (exponent>0);
        long exp = (long)Math.abs((long)exponent); 
        double result = 1;
        double cur = base;
        for(long i=exp;i>0;i=i/2){
            if(i%2==1){
                result = result*cur;
            }
            cur = cur*cur;
        }
        if(!flag){
            result = 1/result;
        }
        return result;
  }
}