June 22, 2022 leetcode daily check-in - the square root of 69.x

1. Topic description and requirements

square root of 69.x

topic description

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

Since the return type is an integer, only the integer part of the result is kept , and the decimal part will be discarded.

Note: Any built-in exponential functions and operators such as pow(x, 0.5) or x ** 0.5 are not allowed .

example

Example 1:

输入：x = 4
输出：2

Example 2:

输入：x = 8
输出：2
解释：8 的算术平方根是 2.82842..., 由于返回类型是整数，小数部分将被舍去。

hint

• 0 <= x <= 231 - 1

2. Problem-solving ideas

General idea:

To find the arithmetic square root of x, you can use the binary search method to find the arithmetic square root of x in [1, x/2+1]. Set two subscripts left and right, the reason for right=x/2+1 is that x/2 is removed, so you need to add 1 to it, and then use left<=right as the loop condition, define mid=(left+right )/2, then compare whether mid is equal to x/mid, if it is, it means that the arithmetic square root of x has been found; mid<x/mid means that the arithmetic square root of x should be on the right, so let left=mid+1; if mid >x/mid, that means the arithmetic square root of x should be on the left, so let right=mid-1. Finally return right.

Specific steps:

1. Set the integer variables left and right as subscripts for auxiliary access
2. Use left<=right as the loop, because if left>right means it is empty
3. Determine between mid and x/mid, enter respectively different judgment statements.
4. Return to right.

3. Specific code

int mySqrt(int x){
    int left=1,right=x/2+1;
    while(left<=right)
    {
        int mid=(left+right)/2;
        if(mid>x/mid)
        {
            right=mid-1;
        }
        else if(mid<x/mid)
        {
            left=mid+1;
        }
        else
        return mid;
    }
    return right;
}