50. Pow(x, n)

Pow(x, n)

Problem Description

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
• n is an integer.
• -104 <= xn <= 104

Approach) Iterative

var myPow = function(x, n) {
    var ans = 1.0;
    var N = Math.abs(n);
    while(N>0)
    {
        if(N%2===1)
        {
            ans *= x;
        }
        x *= x;
        N = Math.floor(N/2);
    }
    return n<0? 1/ ans : ans;
};

Approach) Recursion

var myPow = function(x, n) {
	function helper(x,n){
		if(n === 0) return 1
		if(x === 0) return 0
		const half = helper(x, Math.floor(n/2))

		if(n % 2 === 1){
			return half * half * x
		}else{
			return half * half
		}
	}

	let result = helper(x,Math.abs(n))

	return result = n >= 0 ? result : 1/result
};

OR

var myPow = function(x, n) {
    if(n === 0)
    {
        return 1;
    }
    if(n<0)
    {
        return myPow(1/x, -n);
    }
    return n % 2 === 0 ? myPow(x*x, n/2) : x* myPow(x*x, Math.floor(n/2)); 
};

Conclusion

That’s all folks! In this post, we solved LeetCode problem 50. Pow(x, n)

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Happy coding!