Powx n

Math Recursion

38.3% acceptance

Jan 12, 2026

11575

10520

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

Solution

Rust

Time O(n)

Space O(1)

LeetCode

solution.rs

impl Solution {
  pub fn my_pow(x: f64, n: i32) -> f64 {
    if n == 0 {
      return 1.0;
    }
    
    // Handle negative power
    let (mut base, mut exp) = if n < 0 {
      (1.0 / x, -(n as i64))
    } else {
      (x, n as i64)
    };
    
    let mut result = 1.0;
    
    // Fast exponentiation using binary method
    while exp > 0 {
      if exp % 2 == 1 {
        result *= base;
      }
      base *= base;
      exp /= 2;
    }
    
    result
  }
}