Implement pow(x, n). 下面介绍一下解决该问题的几种方法以及要注意的地方: 1)最直观容易想到的方法就是用递归方法求n个x的乘积,注意考虑n的正负号,时间复杂度为O(n) class Solution { public: double myPow(double x, int n) { ) return 1.0; ) return 1.0/pow(x,-n); ); } }; 2)考虑到n个x相乘式子的对称关系,可以对上述方法进行改进,从而得到一种时间复杂度为O(logn)的方法…

Pow(x, n) Total Accepted: 25273 Total Submissions: 97470My Submissions Implement pow(x, n). 题意:求x的n次幂 思路:二分法 n有可能是负的或正的 当n为负是,pow(x, n) = 1/pow(x, -n) x^n = x^{n/2} * x^{n/2}* x^{n%2} 复杂度:时间O(log n).空间O(1) double power(double x, int n){ if(n == 0) re…

Implement pow(x, n). 刚开始没想到,后来看remlost的博客才写出来,代码很简练: class Solution { public: double pow(double x, int n) { ) /power(x,-n); else return power(x,n); } private: double power(double x, int n){ ) ; ); ==) return v*v; else return v*v*x; } };…

题目: Implement pow(x, n). 链接: http://leetcode.com/problems/powx-n/ 题解: 使用二分法求实数幂,假如不建立临时变量halfPow,直接return计算结果的话会超时,还没仔细研究为什么. Time Complexity - O(logn), Space Complexity - O(1). public class Solution { public double myPow(double x, int n) { if(x == 0…

AC代码 : #include<iostream>#include<cmath>using namespace std;double y;double f(double n){ return 8*pow(n,4)+7*pow(n,3)+2*pow(n,2)+3*n+6;}double find(){ double mid; double a,b; a=0;b=100; while(b-a>1e-6) { mid=(a+b)/2; if(f(mid)<y) a=mid+1…

Implement pow(x, n). 我的做法就比较傻了.排除了所有的特殊情况(而且double一般不可以直接判断==),然后常规情况用循环来做.- -||| 直接用循环,时间复杂度就比较大.应该是用二分法来做.先计算pow(x,n/2).然后再pow(x,n)=pow(x,n/2)*pow(x,n/2) class Solution { public: double power(double x, int n){ ) ; ); == ) return v *v; else return v…

Farmer John is an astounding accounting wizard and has realized he might run out of money to run the farm. He has already calculated and recorded the exact amount of money (1 ≤ moneyi ≤ 10,000) that he will need to spend each day over the next N (1 ≤…

Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is yes, if 6 is a decimal number and 110 is a binary number. Now for any pair of positive integers N​1​​ and N​2​​, your task is to find the radi…

一:用迭代法求 x=√a.求平方根的迭代公式为:X(n+1)=(Xn+a/Xn) /2. #define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #include <stdlib.h> #include <math.h> int main() { double x1, x2; float a; scanf("%f", &a); x2 = 1.0; do { x1 = x2; x2 = (x1 +…

Pow(x, n) Implement pow(x, n). SOLUTION 1: 使用二分法. 1. 负数的情况,使用以下的公式转化为求正数power,另外,考虑到MIN_VALUE可能会造成越界的情况,我们先将负数+1: X^(-n) = X^(n + 1) * XX^n = 1/(x^(-n)) 2. Base case: pow = 0, RESULT = 1; 3. 正数的时候,先求n/2的pow,再两者相乘即可. 当n = -2147483648 必须要特别处理,因为对这个数取反会…