Sqrt(x) 解答

Question

Implement int sqrt(int x).

Compute and return the square root of x.

Solution 1 -- O(log n)

常规做法是用的binary search。注意这里mid一定要是long类型，否则mid * mid会溢出。

 public class Solution {
     public int mySqrt(int x) {
         // Use long instead of int
         long start = 1, end = x, mid;
         while (start + 1 < end) {
             mid = (end - start) / 2 + start;
             long tmp = mid * mid;
             if (tmp == x) {
                 return (int) mid;
             } else if (tmp < x) {
                 start = mid;
             } else {
                 end = mid;
             }
         }
         if (end * end <= x)
             return (int) end;
         return (int) start;
     }
 }

Solution 2 -- Constant Time

我们可以通过以下数学公式缩小问题规模：

因此，问题转换成如何求 log₂N

注意到对于任何数，它的binary representation可以表示为Σ2ⁱ

所以 log₂N 的取值范围为[i, i + 1] i 为最高位的位数。

因此我们就缩小了二分查找的范围。

 public class Solution {
     public int mySqrt(int x) {
         int num = x, digit = 0;
         while (num > 0) {
             num = num >> 1;
             digit++;
         }
         int candidate1 = (int) Math.pow(2, 0.5 * digit);
         int candidate2 = (int) Math.pow(2, 0.5 * (digit - 1));
         long start = candidate2, end = candidate1, mid;
         while (start + 1 < end) {
             mid = (end - start) / 2 + start;
             long tmp = mid * mid;
             if (tmp == x)
                 return (int) mid;
             if (tmp < x)
                 start = mid;
             else
                 end = mid;
         }
         if (end * end <= x)
             return (int) end;
         return (int) start;
     }
 }