README_EN.md

69. Sqrt(x)

中文文档

Description

Given a non-negative integer x, compute and return the square root of x.

Since the return type is an integer, the decimal digits are truncated, and only the integer part of the result is returned.

 

Example 1:

Input: x = 4
Output: 2

Example 2:

Input: x = 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since the decimal part is truncated, 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;
    }
}

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