題目

給定一個double型別的浮點數base和int型別的整數exponent。求base的exponent次方。

思路1

直接使用累乘

    /**
     * 使用累乘
     */
    public double Power(double base, int exponent) {
        double result = 1;
        if (exponent < 0) {
            BigDecimal num = new BigDecimal(base);
            if (num.compareTo(BigDecimal.ZERO)
 
 == 0)
                throw new RuntimeException("分母不能為0");
            for (int i = 0; i < -exponent; i++) {
                result = result * base;
            }
            return 1 / result;
        } else {
            for (int i = 0; i < exponent; i++) {
                result = result * base;
 

            }
            return result;
        }
    }

思路2

二分求解

package com.zhumq.leetcode;
import org.junit.Test;

public class BaseAndExponent {
	public double power(double base,int exponent) throws Exception {
		double result = 1;
		//底數為0時丟擲異常
		if(base == 1) {
			throw new Exception("底數不能為0");
 

		}else if(exponent == 0) {
			//指數為0時值為1
			return 1;
		}
		//指數統一為正
		int n = exponent > 0 ? exponent : (-exponent);
        //右移一位除2
		result = power(base, n >> 1);
        result *= result;
        //與運算判斷指數奇偶性,奇數還需乘上一個base
        if((n &0x1) == 1){
            result *= base;
        }
        //根據指數正負確定返回值
        return exponent > 0 ? result : 1/result;
	}
	
	@Test 
	public void test1() throws Exception {
		double base = 12.0;
		int ex = 3;
		System.out.println(power(base, ex));
	}
}