LeetCode : Sqrt(x)

试题：
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.
代码：
由于不能使用sqrt，只能在1到x之间遍历寻找开方值。对于有序数组使用折半查找。这里注意溢出问题。
如果中点平方刚好等于x，则直接返回；如果一直没找到等于，那么最后只会出现一种情况，那就是找到了low找到结果的ceil，而high找到的结果的floor。

import java.math.*;
class Solution {
    public int mySqrt(int x) {
        if(x<=1) return x;
        int low=1, high=x;
        while(low<=high){
            int mid = low+(high-low)/2;
            int pw = x/mid;
            if(pw==mid){
                return mid;
            }else if(pw<mid){
                high = mid-1;
            }else{
                low = mid+1;
            }
        }
        return high;
    }
}