69. Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

可以用binary search，适合用upperbound的模板，模板：Binary Search - 二分搜索。因为num除了0、1外，sqrt的结果都会<=num/2，这样把start设成0，end设成n/2+2，条件设成mid>x/mid。注意这里不要用mid*mid>x，因为mid*mid会溢出。代码如下：

public class Solution {    public int mySqrt(int x) {        int start = 0, end = x / 2 + 2;        while (start + 1 < end) {            int mid = start + (end - start) / 2;            //don't use mid*mid > x, it will overflow            if (mid > x / mid) {                end = mid;            } else {                start = mid;            }        }        return end - 1;    }}