LeetCode | 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.

 

代码：

class Solution { public: int mySqrt(int x) { if (x < 2) return x; int low = 1, high = x / 2, mid = 1; long long pre = 0, res; while (low <= high) { mid = (low + high) / 2; res = mid; res *= res; if (res == pre || res == x) break; if (res < x) low = mid+1; else high = mid-1; pre = res; } if (pre > x && res != x) return mid - 1; return mid; } };