“【Leetcode】50-56"
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Pow(x,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]
方法一:暴力求解
class Solution {
public:
double myPow(double x, int n) {
long long N = n;
if(N < 0){
x = 1/x;
N = -N;
}
double ans = 1;
for(int i = 0;i < N;i++){
ans *= x;
}
return ans;
}
};
c++ solution:
- 时间复杂度:O(n)
- 空间复杂度:O(1)