Leetcode50.Pow(x, n)

Title Description

Achieve pow (x, n), i.e., a function of n-th power of x.

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
explained: $2^{-2}$ = $1/2^{2}$ = 1/4 = 0.25

Description:

• -100.0 < x < 100.0
• n is a 32-bit signed integer, which is the numerical range [-231 231--1].

answer

Violence (java)

Thinking feeling to do so, then it is entirely up to find a variety of special circumstances. And algorithm time out. . Down is to plan a complete solution to a problem (x
Note:

//对x 和 n 的值进行特殊处理
        //最小值：Integer.MIN_VALUE= -2147483648 （-2的31次方）
        //最大值：Integer.MAX_VALUE= 2147483647  （2的31次方-1）
        //这里需要使用long类型，因为如果传入的n = -2147483648； 那么转成正数就丢失，所以要使用long
        long N = n;
        if(N < 0){
            N = -N;
            x = 1/x;
        }
        return  pow(x,N);

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

Complexity Analysis

• time complexity: $O (N)$
• Space complexity: $O (1)$

Divide and Conquer (java)

Idea: with the idea of ​​partition, the calculation requires a step by step two. Note that the last of the merger.

class Solution {
    public double ans(double x, int n) {
        if (n == 0) return 1.0;
        double result = ans(x, n / 2);
        if (n % 2 == 0) {
            return result * result;
        } else {
            return result * result * x;
        }
    }
    public double myPow(double x, int n) {
        if (n >= 0) return ans(x, n);
        double result;
        int N = - n;
        result = ans(x, N);
        result = 1.0 / result;
        return result;
    }
}

• time complexity: $O(\log N)$
• Space complexity: $O(\log N)$