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.
 


題意：開方
這裏肯定是不使用Math庫的前提下，
我們可以利用二分的思想去開方，開方反過來就是平方，Integer.MAX_VALUE 開方是46340，也就是最大的解，因為存在越界的情況
這個數字需要我們提前算出來。之後只要二分去找，mid*mid <= target , (mid+1)*(mid+1) > target 就行了

class Solution {
    public int mySqrt(int x) {
        int l = 0;
        int r = 46340;
        if (r*r < x)
            
 
return r;
        if (x == 1 || x == 0)
            return x;
        while (r - l > 5) {
            int mid = (l + r) / 2;
            if (mid * mid < x) {
                l = mid;
            }
            else if (mid * mid > x) {
                r = mid;
            }
            
 
else
                return mid;
        }
        for (int i = l; i <= r; i++) {
            if (i*i == x)
                return i;
            if (i*i < x && (i+1)*(i+1) > x)
                return i;
        }
        return 0;
    }
}

[LeetCode] 69. Sqrt(x)