LeetCode Top Interview Questions 50. Pow(x, n) (Java版; Medium)

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LeetCode Top Interview Questions 50. Pow(x, n) (Java版; Medium)

题目描述

Implement pow(x, n), which calculates x raised to the power n (x^n).

Example 1:

Input: 2.00000, 10
Output: 1024.00000
Example 2:

Input: 2.10000, 3
Output: 9.26100
Example 3:

Input: 2.00000, -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25

第一次做; 参考了LeetCode的最优解; 这个算法也叫快速幂法

/*
核心:2^16 == 4^8
和不使用加减乘除相加有点像
*/
class Solution {
    public double myPow(double x, int n) {
        //base case
        if(x==0)
            return 0;
        if(n==0)
            return 1;
        //自顶向下; 本题最麻烦的地方在于n=Integer.MIN_VALUE
        double res = 0;
        if(n<0){
            if(n==Integer.MIN_VALUE){
                n++;
                n = -n;
                x = 1/x;
                return x*x*myPow(x*x, n/2);
            }
            else{
                n = -n;
                x = 1/x;
                // return n%2==0?myPow(x*x, n/2) : x*myPow(x*x, n/2);
            }
        }
        return n%2==0?myPow(x*x, n/2) : x*myPow(x*x,n/2); 
    }
}

第一次做; 分治; 自顶向下; 使用memo数组记录计算过的幂, 但是内存超出限制,291 / 304; 绊脚案例:x=0.00001, n=2147483647; 看来需要自底向上?

class Solution {
    public double myPow(double x, int n) {
        double res=1;
        if(x==0)
            return 0;
        else{
            if(n==0)
                return 1.0;
            //分治; 自顶向下
            int tmp = Math.abs(n);
            //2147483647
            double[] memo = new double[tmp];
            res = core(memo, x, tmp);
        }
        return n>0?res : 1/res;
    }
    public double core(double[] memo, double x, int n){
        //base case
        if(n==0)
            return 1;
        if(n==1)
            return x;
        if(memo[n-1]!=0)
            return memo[n-1];
        //
        double leftRes = core(memo, x, n/2);
        double rightRes = core(memo, x, n-n/2);
        memo[n-1] = leftRes*rightRes;
        return memo[n-1];
    }
}

第一次做; 循环版竟然超时; 291 / 304; 看样子得改成分治法

/*
求幂
按照底数是否为0分成两种情况讨论
*/
class Solution {
    public double myPow(double x, int n) {
        double res=1;
        if(x==0)
            return 0;
        else{
            if(n==0)
                return 1.0;
            for(int i=0, m=Math.abs(n); i<m; i++){
                res = res * x;
            }
        }
        return n>0?res : 1/res;
    }
}

快速幂算法; 自底向上; 来自力扣题解

class Solution {
    public double myPow(double x, int n) {
        long N = n;
        if (N < 0) {
            x = 1 / x;
            N = -N;
        }
        double ans = 1;
        double current_product = x;
        for (long i = N; i > 0; i /= 2) {
            if ((i % 2) == 1) {
                ans = ans * current_product;
            }
            current_product = current_product * current_product;
        }
        return ans;
    }
};