(二分查找 拓展) leetcode 69. Sqrt(x)

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.--------------------------------------------------------------------------------------------这个问题，其实很简单（用sqrt直接OK了），不过，我用二分查找解决这个题C++代码： class Solution { public: int mySqrt(int x) { if(x <= 1) return x; int left = 0; int right = x; while(left <= right){ int mid = left + (right - left)/2; if(x / mid >= mid){ left = mid + 1; } else right = mid - 1; } return right; } };

 用sqrt：

class Solution { public: int mySqrt(int x) { if(x <= 1)return x; return (int)(sqrt(x)); } };

 

转载于:https://www.cnblogs.com/Weixu-Liu/p/10758509.html

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