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Leetcode050-pow(x,n)

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Solution Report of LeetCode Acceptted

Description

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

Example

Example 1:

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Input: 2.00000, 10
Output: 1024.00000

Example 2:
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Input: 2.10000, 3
Output: 9.26100

Example 3:
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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 [$−2^{31}$, $2^{31} − 1$]

Solution

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class Solution {
    public double myPow(double x, int n) {
        if (n == 0) return 1;
        
        if (n == Integer.MIN_VALUE){
            x *= x;
            n /= 2;
        }//beacuse Integer.MAX_VALUE = abs(Integer.MIN_VALUE)-1
        if (n < 0){
            n = -n;
            x = 1/x;
        }
        return (n % 2 == 0) ? myPow(x * x, n / 2) : x * myPow(x * x, n / 2); 
    }
}