【leetcode】69. Sqrt(x)（二分）

1、题目描述

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

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: 4
Output: 2
Example 2:

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

2、问题描述：

• 实现求平方根的操作。
• 如果是小数，返回整数部分。

3、问题关键：

• 二分法，依次逼近。

4、C++代码：

class Solution {
public:
    int mySqrt(int x) {
        int l = 0, r = x;
        while(l < r) {
            int mid = l + 1ll + r >> 1;
            if (mid <= x / mid) l = mid;//这个地方注意是<=，不要忘记等号。
            else r = mid - 1;
        }
        return l;
    }
};