Description
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.
Solution
Binary search, time O(logn), space O(1)
用二分解就可以了，注意二分判断的条件
if (mid < x / mid)
这里一定要保证mid > 0，否则会抛Runtime Exception，这就要求对x < 2进行预处理。
也可以用一个long square = (long) mid * mid，然后用square和x作比较来二分，也比较简单。
注意如果x不是perfect square，需要返回right而非left！
class Solution {
public int mySqrt(int x) {
if (x < 0) {
return -1;
}
if (x < 2) {
return x;
}
int left = 0;
int right = x;
while (left <= right) {
int mid = left + (right - left) / 2;
if (mid < x / mid) { // ensure mid > 0!
left = mid + 1;
} else if (mid > x / mid) {
right = mid - 1;
} else {
return mid; // x is perfect square
}
}
return right; // x is not a perfect square, round down
}
}
或者用long表示pow也可以，写起来更方便：
class Solution {
public int mySqrt(int x) {
int left = 0;
int right = x;
while (left <= right) {
int mid = left + (right - left) / 2;
long pow = (long) mid * mid;
if (pow > x) {
right = mid - 1;
} else {
left = mid + 1;
}
}
return right;
}
}
