[LeetCode]实现数学计算

乘方

思路是：pow(x,n) = pow(x,n/2)*pow(x,n-n/2)

递归实现

public double myPow(double x, int n) {
        if (n==0) return 1.0;
        else if (n>0)
        {
            double half = (double) myPow(x,n/2);
            if (n%2==0) return half*half;
            else return half*half*x;
        }
        else {
            if (n==Integer.MIN_VALUE)
            {
                double half = myPow(x,n/2);
                return (half*half);
            }
            n=-n;
            double half = myPow(x,n/2);
            if (n%2==0) return 1.0/(half*half);
            else return 1.0/(half*half*x);
        }
    }

开方

二分法或者牛顿法

public int mySqrt(int x) {
        /*
        二分法一个一个寻找答案
         */
        if (x==0) return 0;
        long sta = 1;
        long end = x/2+1;
        while (sta<end)
        {
            long mid  = (sta+end)/2;
            long cur = mid*mid;
            if (cur==x) return (int)mid;
            if (cur>x) end = mid-1;
            else sta = mid+1;
        }
        return (int)sta;
    }