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
Input: 4
Output: 2
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since
the decimal part is truncated, 2 is returned.
Thoughts
- 牛顿法： \(x_{n+1} = x_n - \frac{f(x_n)}{f^{'}(x_n)} \tag{1}\)
- 二分法：
- 二分搜索法，从
[1,x/2]的区间开始搜索，每次减半，最后发现结果Code(JAVA)
- 二分搜索法，从
//newton method:
public int mySqrt(int x) {
if(x == 1)
return x;
double guess = x/2;
while(Math.abs(guess * guess - x) > 0.1) {
guess = guess / 2 + x / (2 * guess) ;
}
return (int)guess;
}
//binary search
public int mySqrt(int x) {
if(x == 0 || x == 1)
return x;
int low = 1, high = x / 2;
while(low <= high) {
int mid = (low + high) / 2;
long squre = (long)mid * mid;
if(squre == x)
return mid;
else if(squre < x)
low = mid + 1;
else
high = mid - 1;
}
return high;
}
