LeetCode 50. Pow(x, n)

原题链接在这里：https://leetcode.com/problems/powx-n/

题目：

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

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

Note:

• -100.0 < x < 100.0
• n is a 32-bit signed integer, within the range [−231, 231 − 1]

题解：

建立一个helper function返回n为正数时的power, 主函数里分开调用即可。

corner case 当n = Integer.MIN_VALUE, -n会overflow. 所以helper function 用long型.

helper中采用递归，原理就是x^n = x^(n/2) * x^(n/2) * (n == 奇数)*x, 终止条件是n==0时返回一.

Note: calculate half = powPositive(x, n / 2) first to cache extra result.

Time Complexity: O(logn).

Space: O(logn), stack space.

AC Java:

class Solution {
    public double myPow(double x, int n) {
        if(n<0){
            return 1/powPositive(x, -1*(long)n);
        }
        
        return powPositive(x, n);
    }
    
    private double powPositive(double x, long n){
        if(n == 0){
            return 1.0;
        }
        
        double half = powPositive(x, n/2);
        if(n%2 == 1){
            return half * half * x;
        }else{
            return half * half;
        }
    }
}

Iterative 方法.

Time Complexity: O(logn).

Space: O(1).

AC Java:

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

这道题和Sqrt(x)以及Divide Two Integers都是原有公式的题。这类题目一般用二分法.