LeetCode69. Sqrt(x)

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;
    }
};
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