README_EN.md

69. Sqrt(x)

中文文档

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

Solutions

Binary search.

Python3

class Solution:
    def mySqrt(self, x: int) -> int:
        left, right = 0, x
        while left < right:
            mid = (left + right + 1) >> 1
            # mid*mid <= x
            if mid <= x // mid:
                left = mid
            else:
                right = mid - 1
        return left

Java

class Solution {
    public int mySqrt(int x) {
        int left = 0, right = x;
        while (left < right) {
            int mid = (left + right + 1) >>> 1;
            if (mid <= x / mid) {
                // mid*mid <= x
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }
}

C++

class Solution {
public:
    int mySqrt(int x) {
        long long left = 0, right = x;
        while (left < right) {
            long long mid = left + ((right - left + 1) >> 1);
            if (mid <= x / mid)
                left = mid;
            else
                right = mid - 1;
        }
        return (int)left;
    }
};

Go

func mySqrt(x int) int {
	left, right := 0, x
	for left < right {
		mid := left + (right-left+1)>>1
		if mid <= x/mid {
			left = mid
		} else {
			right = mid - 1
		}
	}
	return left
}

JavaScript

/**
 * @param {number} x
 * @return {number}
 */
var mySqrt = function (x) {
    let left = 0;
    let right = x;
    while (left < right) {
        const mid = (left + right + 1) >>> 1;
        if (mid <= x / mid) {
            left = mid;
        } else {
            right = mid - 1;
        }
    }
    return left;
};

C#

public class Solution {
    public int MySqrt(int x) {
        int left = 0, right = x;
        while (left < right)
        {
            int mid = left + right + 1 >> 1;
            if (mid <= x / mid)
            {
                left = mid;
            }
            else
            {
                right = mid - 1;
            }
        }
        return left;
    }
}

Rust

impl Solution {
    pub fn my_sqrt(x: i32) -> i32 {
        if x < 2 {
            return x;
        }
        let mut l = 1;
        let mut r = x / 2;
        while l < r {
            let mid = (l + r + 1) >> 1;
            if x / mid < mid {
                r = mid - 1
            } else {
                l = mid;
            }
        }
        l
    }
}

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