lintcode:Pow(x, n)

Pow(x, n)

Implement pow(x, n).

解题

直接顺序求解，时间复杂度O(N)

public class Solution {
    /**
     * @param x the base number
     * @param n the power number
     * @return the result
     */
    public double myPow(double x, int n) {
        // Write your code here
        if(x == 0)
            return 0;
        if(n == 0)
            return 1.0;
        if( n<0)
            return 1.0/(myPow(x,-n));
        double res = x;
        while(n>1){
            res*=x;
            n--;
        }
        return res;
    }
}

class Solution:
    # @param {double} x the base number
    # @param {int} n the power number
    # @return {double} the result
    def myPow(self, x, n):
        # Write your code here
        return x**n

 递归方式

对n时候奇数还是偶数的时候进行讨论

public class Solution {
    /**
     * @param x the base number
     * @param n the power number
     * @return the result
     */
    public double myPow(double x, int n) {
        // Write your code here
        if(x == 0)
            return 0;
        if(n == 0)
            return 1.0;
        if( n<0)
            return 1.0/(myPow(x,-n));
        double res = myPow(x,n/2);
        
        if(n%2==1){
            res = res * res *x;
        }else{
            res = res * res;
        } 
        return res;
    }
}

递归程序中间需要比较多的栈，容易发生内存溢出的情况

 改写成循环的形式，效果更好，参考于libsvm源码

public class Solution {
    public double myPow(double x, int n) {
        if(n==0)
            return 1.0;
        if(x==1)
            return 1.0;
        if(n<0){
            if (n == Integer.MIN_VALUE)
                return myPow(x, n+1)/x;
            else 
                return 1.0/myPow(x, -n);
        }
            
        
        double tmp = x;
        double ret = 1.0;
        for(int time = n;time>0;time/=2){
            if(time%2==1)
                ret *= tmp;
            tmp*=tmp;
        }
        return ret;
    }
}

上面程序对n是最小值得时候进行了处理，LeetCode测试样例有 2 ，Integer.MIN_VALUE ,上面单独处理，可以通过