LeetCode in TypeScript

50. Pow(x, n)

Medium

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

Solution

function myPow(x: number, n: number): number {
    let nn = BigInt(n)
    let res = 1.0
    if (n < 0) {
        nn = -nn
    }
    while (nn > 0) {
        if (nn % 2n === 1n) {
            nn--
            res *= x
        } else {
            x *= x
            nn /= 2n
        }
    }
    return n < 0 ? 1.0 / res : res
}

export { myPow }

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