# 69. Sqrt(x)

[Link](https://leetcode.com/problems/sqrtx/)

Implement `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.
```

**Solution**

處理數相關的問題，盡量透過減法、除法、shift位元的方式進行計算，減少溢位出錯

此題用binary Search求解。 narrow down可用的range直到left == right為止。

此時left/right值可能跟所求一步之遙，要看上次binary search的運作是判斷到left導致left增加 or right導致right減少。\
所以還要額外判斷哪個才是對的。

```
 public int mySqrt(int x) {
        if (x == 0) return 0;
        int right = x, left = 1;
        while(right > left){
            int mid = left + (right - left)/2;
            if(mid == x/mid) return mid;
            else if (mid > x/mid)
            {
                right = mid - 1;
            }else{
                left = mid+1; 
            }
        }
        return (left > x/left) ? left-1 : left;
    }
```
