lintcode: Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

Example
sqrt(3) = 1

sqrt(4) = 2

sqrt(5) = 2

sqrt(10) = 3

Challenge
O(log(x))

求非线性方程的解可以使用二分法。当然前提是该方程所表示的函数在一定区间内与x的交点只有一个。

x√

x√<=(x+1)/2

class Solution {
public:
    /**
    * @param x: An integer
    * @return: The sqrt of x
    */
    int sqrt(int x) {
        // write your code here
        if (x < 0){
            return -1;
        }

        long long l = 0, r = x/2+1;


        while (l<=r){

            long long mid = l + (r - l) / 2;
            long long tmp = mid*mid - x;

            if ( tmp== 0){
                return mid;
            }
            else if (tmp>0){
                r = mid-1;
            }
            else{
                l = mid+1;
            }
        }
        return r;//return l出错

    }
};

还有牛顿迭代法可以求平方根