題目描述

• 實(shí)現(xiàn)函數(shù) double Power(double base, int exponent)蒿讥，求 base 的 exponent 次方
• 不得使用庫(kù)函數(shù)，同時(shí)不需要考慮大數(shù)問(wèn)題

題目解讀

• 劍指Offer 110

代碼

• 思路一咒林、自以為題目簡(jiǎn)單的解法

#include<iostream>
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<<ss.Power(2, 3)<<endl;
}

• 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;
    }
};

• 和思路三思路一樣描扯，但是細(xì)節(jié)方面處理更加優(yōu)越。判斷偶數(shù)/奇數(shù)趟薄。除二運(yùn)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;
        }
        
        // 偶數(shù)
        if(exponent&1 == 0){
            result = core(base, exponent);
        }
        else{
            result = base * core(base, exponent-1);
        }
        
        if(flag == true){
            result = 1.0 / result;
        }
        return result;
    }
};

總結(jié)展望

• 三種思路看書(shū)即可
• 很好的題目绽诚，想想自己只能想到第一種解法，真是需要多加練習(xí)