LeetCode 50. Pow(x, n)

[woodyiiiiiii]

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

Example 1:

Input: x = 2.00000, n = 10
Output: 1024.00000

Example 2:

Input: x = 2.10000, n = 3
Output: 9.26100

Example 3:

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

Constraints:

• -100.0 < x < 100.0
• -231 <= n <= 231-1
• -104 <= xn <= 104

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题目要求我们重写pow方法。

对幂次方n做一个二分，递归调用。

class Solution {
    public double myPow(double x, int n) {
        double temp; 
        if(n == 0) return 1; 
        temp = myPow(x, n/2);
        if (n % 2 == 0) 
            return temp*temp; 
        else{ 
            if(n > 0) 
                return x * temp * temp; 
            else
                return (temp * temp) / x; 
        } 
    }
}

或者使用快速幂。注意，因为-2147483648无法用Math.abs()化为正数，所以要特殊处理。

class Solution {
    public double myPow(double x, int n) {
        if (n == 0) return 1.0;
        if (n == Integer.MIN_VALUE) {
            if (x == 1.0) return 1.0;
            if (x == -1.0) return 1.0;
            return 0.0;
        }
        int absN = Math.abs(n);
        double base = x, ans = 1.0;
        while (absN > 0) {
            if (absN % 2 == 1) {
                ans *= base;
            }
            base *= base;
            absN >>= 1;
        }
        return n < 0 ? 1 / ans : ans;
    }
}

类似题目：

• 2550. Count Collisions of Monkeys on a Polygon

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参考资料：

• https://leetcode.com/problems/powx-n/