[算法作业][LeetCode] 50. Pow(x, n) -- 分治法

[算法作业][LeetCode] 50. Pow(x, n) – 分治法

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Implement pow(x, n).

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对于 xn ，可以通过分治的方法，设 n=2k ：

xn=xn/2∗xn/2∗xn%2
xn/2=xn/4∗xn/4∗xn/2%2
.
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xn/2k−1=x1∗x1∗xn/2k−1%2

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function pow(x, n):
  input x, n
  output xn
    if (n == 0) return 1;
    return pow(x, n/2) * pow(x, n/2) * pow(x%2)

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function pow(x, n):
  input x, n
  output xn
    res = 1; base = x;
    while (n):
        if (n % 2) res *= base;
        base *= base;
        n /= 2;
    return res;

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递归

class Solution {
public:
    double myPow(double x, int n) {
        if (n == 0) return 1;
        double tmp = myPow(x, n/2);
        double res = tmp * tmp;
        if (n > 0) {
            if (n%2) return res * x;
            else return res;
        } else {
            if (n%2) return res / x;
            else return res;
        }
    }
};

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非递归

double myPow(double x, int n) {
        double res = 1;
        double base = x;
        while (n) {
            if (n%2) {
                if (n < 0) res /= base;
                else res *= base;
            }
            base *= base;
            n /= 2;
        }
        return res;
    }