一、题目
Given a non-negative integer x, return the square root of x rounded down to the nearest integer. The returned integer should be non-negative as well.
You must not use any built-in exponent function or operator.
For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python.
Example 1:
Input: x = 4
Output: 2
Explanation: The square root of 4 is 2, so we return 2.
Example 2:
Input: x = 8
Output: 2
Explanation: The square root of 8 is 2.82842…, and since we round it down to the nearest integer, 2 is returned.
Constraints:
0 <= x <= 231 - 1
二、题解
class Solution {
public:
int mySqrt(int x) {
int l = 0,r = x,res = 0;
while(l <= r){
int mid = l + ((r-l) >> 1);
if((long long)mid * mid <= x){
res = mid;
l = mid + 1;
}
else{
r = mid - 1;
}
}
return res;
}
};