(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.
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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)); } };