# 50. Pow(x, n) (Medium)

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

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 = 1/22 = 1/4 = 0.25
```

Note:

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

## Solutions

``````class Solution {
// keep results already calculated, avoid repetitive computation
Map<String, Double> cache = new HashMap<>();

public double myPow(double x, int n) {
if (n == 0) {
return 1d;
}

if (x == 1) {
return x;
}

// Use long type instead of int such that avoid overflow issue
long newN = n;
if (newN < 0) {
x = 1 / x;
newN = -newN;
}

return solve(x, newN);
}

public double solve(double x, long n) {
if (n == 0) {
return 1;
}

if (n == 1) {
return x;
}

String key = x + "#" + n;
if (cache.containsKey(key)) {
return cache.get(key);
}

double val = solve(x, n / 2);

// if n is even
double ans = val * val;

// if n is odd, do not forget there is still 1 x not multiplied.
if (n % 2 == 1) {
ans = val * val * x;
}

cache.put(key, ans);

return ans;
}
}
``````