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.
此方法為二元逼近法，這題會超過 int 大小所以變數記得要宣告 long 最後回傳時再強制轉型。
- 二元搜尋
- Run Time: 12 ms
- 時間複雜度: O(log n)
- 空間複雜度: O(1)
int mySqrt(int x) {
long r=x;
while(r*r>x){
r = (r + x / r) / 2;
}
return (int)r;
}