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
- 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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- 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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- 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.