Implement int sqrt(int x).
Compute and return the square root of x.
x is guaranteed to be a non-negative integer.
Example 1:
Input: 4
Output: 2
Example 2:
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since we want to return an integer, the decimal part will be truncated.
题目标签：Math
题目给了我们一个x， 让我们找到平方根。
可以利用binary search 来做，首先找到一个范围，不需要是从0 到 x。因为 sqrt(x) 一定是小于等于 x / 2 + 1的。
所以起始范围为 0 到 x / 2 + 1;
还要注意 sq = mid * mid。 这里要检查 overflow。
Java Solution:
Runtime beats 53.10%
完成日期：06/12/2017
关键词：Binary Search
关键点：检查overflow
class Solution
{
public int mySqrt(int x)
{
int left = 0;
int right = x / 2 + 1; while(left <= right)
{
int mid = left + (right - left)/2;
int sq = mid * mid; if(mid != 0 && sq / mid != mid) // check overflow
{
right = mid - 1; // meaning mid is too big, go to left part
continue;
} if(sq == x)
return mid;
else if(sq < x)
left = mid + 1;
else
right = mid - 1; } return right;
}
}
参考资料：N/A
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