69. Sqrt(x)

题目：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.

思路是：二分法查找平方根，如果中心的的平方大于X，说明平方根在左区间，则right = mid - 1，反之left = mid + 1。

class Solution {
    public int mySqrt(int x) {
        if (x == 0)
            return 0;
        int left = 1, right = x;
        while (true) {
            int mid = left + (right - left)/2;
            if (mid > x/mid) {  //不要用mid * mid > x，可能会溢出
                right = mid - 1;
            } else {
                if (mid + 1 > x/(mid + 1))
                return mid;
                left = mid + 1;
            }
         }
    }
}