LeetCode-50.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]

遍历n，时间复杂度为O(n)，但是Time Limit Exceeded 

public double myPow(double x, int n) {//my
        double re = 1;
        for (int i = 1; i <= Math.abs(n); i++) {
            re *=x;
        }
        return n<0?(1/re):re;
    }

使用分治的方法，时间复杂度为O(logn)

class Solution {
    public double myPow(double x, int n) {//分治 递归 my
        long y = n;//一定要是long 否则n=-2147483648时会溢出
        if(y<0){
            y=-y;
        }
        double re = calPow(x,y);
        return n<0?(1/re):re;
    }
    private double calPow(double x,long n){
        if(0==n){
            return 1;
        }
        if(1==n){
            return x;
        }
        double re =calPow(x,n/2);
        re = re*re;
        if (n%2==1){
            re*=x;
        }
        return re;
    }
}

非递归写法 时间复杂度O(logn)

 public double myPow(double x, int n) {
        long y = n;
        if(y<0){
            y=-y;
        }
        double re = 1;
        while(0!=y){
            long bit = 1&y;
            
            if(0!=(1&y)){
                re *= x;
            }
            x =x*x;
            y= y>>1;
        }
        return n<0?(1/re):re;
    }