Sqrt(x)

Description

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

Code

基本概念同 Binary Search 101。

可以用除法避免 integer overflow。

class Solution {
public:
    int mySqrt(int x) {
        if(x == 0) return 0;
        int l = 1, r = x;
 
        while(l < r) {
            int mid = l + (r - l) / 2;
            if(mid == x / mid) return mid;
            else if(mid < x / mid) {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
 
        return l == x / l ? l : l - 1;
    }
};

和 Valid Perfect Square 是同個概念。

class Solution {
public:
    int mySqrt(int x) {
        long long l = 1, r = x;
        while(l < r) {
            long long m = l + (r - l) / 2;
            long long product = m * m;
            if(product < x) {
                l = m + 1;
            } else {
                r = m;
            }
        }
        return l * l > x ? l - 1 : l;
    }
};

Source

• Sqrt(x) - LeetCode