題目描述：

mplement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4
Output: 2

Example 2:

Input:
 
 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

 解題思路：

    x的平方根值應該在區間[1,x]，而且sqrt=x/sqrt，所以用二分查詢sqrt。根據題意有，若x=8,則輸出應該是3，迴圈條件是l<=h，當跳出迴圈時，h比l小1，所以返回值應該是h。具體思路及程式碼如下：

package Leetcode_Github;

public class BinarySearch_SqrtX_69_1113 {
    public int MySqrt(int x) {
        //判斷輸入是否合法
        if (x <= 1) {
            return x;
        }
        int l = 1;
        int h = x;
        while (l <= h) {
            int mid = l + (h - l) / 2;
            int sqrt = x / mid;
            if (mid == sqrt) {
                return mid;
            } else if (mid > sqrt) {
                h = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        return h;
    }
}