69.Sqrt(x)

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Riguz留言 | 贡献2024年2月29日 (四) 14:02的版本 (创建页面,内容为“=Description= {{LeetCode |id=sqrtx |no=69 |difficulty=Easy |category=Math |collection=Top 150 |title=Sqrt(x) |summary=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: <syntaxhighlight lang="bash"> Input: x = 4 Output:…”)
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Description

#69 Sqrt(x) Easy
Math Top 150
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.

Binary search

Runtime 1ms
Memory 40.82MB
class Solution {
    public int mySqrt(int x) {
        int l = 1, r = x;
        while(l <= r){
           int m = (r-l)/2 + l;
           if(m > x/m) r = m-1;
           else if(m < x/m) l = m+1;
           else return m;
        }
        return r;
    }
}