README.md

Pow(x, n)

Description

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 = 1/2^2 = 1/4 = 0.25

Note:

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

Tags: Math, Binary Search

思路

题意是让你计算 x^n，如果直接计算肯定会超时，那么我们可以想到可以使用二分法来降低时间复杂度。

class Solution {
    public double myPow(double x, int n) {
        if (n < 0) return helper(1 / x, -n);
        return helper(x, n);
    }

    private double helper(double x, int n) {
        if (n == 0) return 1;
        if (n == 1) return x;
        double d = helper(x, n >>> 1);
        if (n % 2 == 0) return d * d;
        return d * d * x;
    }
}

结语

如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：awesome-java-leetcode