<剑指Offer>面试题16: 数值的整数次方

题目描述

• 实现函数 double Power(double base, int exponent)，求 base 的 exponent 次方
• 不得使用库函数，同时不需要考虑大数问题

题目解读

• 剑指Offer 110

代码

• 思路一、自以为题目简单的解法

#include
using namespace std;

class Solution {
public:
    double Power(double base, int exponent) {
        double result = 1.0;
        for(int i = 0; i < exponent; i++){
            result = result * base;
        }
        return result;
    }
};

int main(){
    Solution ss;
    cout<

• 思路二、

class Solution {
public:
    double Power(double base, int exponent) {

        if(base == 0) return 0;
        if(exponent == 0) return 1;

        bool flag = true;
        if(exponent < 0){
            flag = false;
            exponent = -1 * exponent;
        }

        double result = 1.0;
        for(int i = 0; i < exponent; i++){
            result = result * base;
        }

        if(flag == false){
            result = 1.0 / result;
        }
        return result;
    }
};

• 思路三、

class Solution {
public:
    double PowerCore(double base, int exponent){
        double result = 0;

        if(exponent == 1){
            return base;
        }
        else{
            result = PowerCore(base, exponent/2);
            return result * result;   
        }
    }

    double Power(double base, int exponent) {
        double result = 1.0;
        bool flag = true;

        if(base == 0) return 0;
        if(exponent == 0) return 1;

        if(exponent < 0){
            flag = false;
            exponent = -1 * exponent;
        }

        if(exponent/2 == 0){
            result = PowerCore(base, exponent);
        }
        else{
            result = PowerCore(base, exponent - 1);
            result = result * base;
        }
        
        if(flag == false){
            result = 1.0 / result;
        }
        return result;
    }
};

• 和思路三思路一样，但是细节方面处理更加优越。判断偶数/奇数。除二运算

class Solution {
public:
    double core(double base, int exponent){
        if(exponent == 1){
            return base;
        }
        else{
            double tt = core(base, exponent>>1);
            return tt * tt;
        }
    }
    
    double Power(double base, int exponent) {
        if(base == 0) return 0;
        if(exponent == 0) return 1;
        double result = 1.0;
        
        bool flag = false;
        if(exponent < 0){
            flag = true;
            exponent *= -1;
        }
        
        // 偶数
        if(exponent&1 == 0){
            result = core(base, exponent);
        }
        else{
            result = base * core(base, exponent-1);
        }
        
        if(flag == true){
            result = 1.0 / result;
        }
        return result;
    }
};

总结展望

• 三种思路看书即可
• 很好的题目，想想自己只能想到第一种解法，真是需要多加练习