Offer subject to prove safety 11: Implement function double Power (double base, int exponent), seeking exponent of the power base. (Java)

Interview questions 11 : Given a double base type and floating-point type int integer exponent. Seeking exponent of the power base.

Problem-solving ideas

思路1：brute force 累乘法，时间复杂度O(n)

挨个乘，exponent 为 n，则累乘 n 次得出结果。

思路2：使用递归，时间复杂度O(logn)

当n为偶数，a^n =（a^n/2）*（a^n/2）

当n为奇数，a^n = a^[(n-1)/2] * a^[(n-1)/2] * a

思路3：同递归的思路，但采用循环法（递归和循环都是可以互换的）

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Code

1. brute force multiplication tired

    public double Power(double base, int exponent) {
        if(base == 0) {
            return 0;
        }
        double result = 1.0;
        int positiveExponent = Math.abs(exponent);
        for(int i=1; i<=positiveExponent; i++) {
            result *= base;
        }
        return exponent < 0 ? 1/result : result;
  }
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2. Recursion

public class Solution {
    public double Power(double base, int exponent) {
        if(base == 0) {
            return 0;
        }
        int positiveExponent = Math.abs(exponent);
        double result = PowerPositiveExponent(base, positiveExponent);
        return exponent < 0 ? 1 / result : result;
    }
    
    private double PowerPositiveExponent(double base, int n){
        if(n == 0) {
            return 1.0;
        }
        if(n == 1) {
            return base;
        }
        double result = PowerPositiveExponent(base, n >> 1);
        result *= result;
        if((n & 1) == 1) {
            result *= base;
        }
        return result;
    }
}
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3. Round-robin

public class Solution {
    public double Power(double base, int exponent) {
        if(base == 0) {
            return 0;
        }
        double result = 1.0;
        int positiveExp = Math.abs(exponent);
        while(positiveExp != 0){
            //此处判断奇数，这一步至少会走两次：
            //如果是奇数，第一次就会进入这个判断。移位到最后为1时，也会进入这个判断
            if((positiveExp & 1) == 1){
                result *= base;
            }
            base *= base;
            positiveExp = positiveExp >> 1;
        }
        return exponent < 0 ? 1 / result : result;
    }
}
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to sum up

Generally think of the idea behind, remember you can. Idea is to save intermediate values, to avoid double counting.

On this idea of the nature and " Fibonacci algorithm ideological hold intermediate values" are the same.