【剑指offer】Java数值的整数次方

it2022-05-05  269

github地址:https://github.com/PLLLLLLL/offer-Java

/** * 【题目描述】 * 给定一个double类型的浮点数base和int类型的整数exponent。求base的exponent次方。 * 考虑多种情况,注意代码完整性 */ // ======常规运算====== public class Solution { public double Power(double base, int exponent) { double result = 0.0; if (base == 0 && exponent == 0) return -1; if (exponent<0) { result = PowerWithUnsignedExponent(base,-exponent); result = 1.0/result; } else { result = PowerWithUnsignedExponent(base,exponent); } return result; } public double PowerWithUnsignedExponent(double base, int exponent) { // 注意 double double result = 1.0; for (int i=1;i<=exponent;i++){ // 连乘 result = base*result; } return result; } } // ======效率更高的位运算====== public class Solution { public double Power(double base, int exponent) { double result = 0.0; if (base == 0 && exponent == 0) return -1; if (exponent<0) { result = PowerWithUnsignedExponent(base,-exponent); result = 1.0/result; } else { result = PowerWithUnsignedExponent(base,exponent); } return result; } public double PowerWithUnsignedExponent(double base, int exponent) { // 如果指数为0,返回1 if (exponent == 0) { return 1; } // 指数为1,返回底数 if (exponent == 1) { return base; } // 递归求一半的值 double result = PowerWithUnsignedExponent(base, exponent >> 1); // 求最终的值,如果是奇数就还要剩以一次底数 result *= result; if (exponent % 2 != 0) { result *= base; } // 返回结果 return result; } }

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