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.
题意:
求一个数的平方根,保证这个数大于等于0.且如果是小数那么就舍掉取整数。
思路:
观察可知,题目要求的,除了1,其余的结果都小于等于x/2.可以采用二分法,在0-x/2之间试探找出答案,降低时间复杂度。
代码:
public class Solution {
public int MySqrt(int x) {
double b=0,e=x;
double r=1,mid=1;
while(Math.Abs(r-x)>0.00001)
{
mid=(e+b)/2;
r=mid*mid;
if(r>x)
e=mid;
else
b=mid;
}
return (int)mid;
}
}