[lintcode]141. x的平方根

原创已于 2024-05-16 20:46:29 修改 · 265 阅读

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#蓝桥杯 #linq #c++

于 2018-04-23 13:48:34 首次发布

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本文介绍了一种计算整数平方根的有效算法。通过使用二分查找法，可以在O(logn)的时间复杂度内找到任意非负整数x的平方根的整数部分。文章提供了两种不同的C++实现方式。

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链接：https://www.lintcode.com/zh-cn/problem/sqrtx/

实现 int sqrt(int x) 函数，计算并返回 x 的平方根。

您在真实的面试中是否遇到过这个题？

Yes

样例

sqrt(3) = 1

sqrt(4) = 2

sqrt(5) = 2

sqrt(10) = 3

class Solution {
public:
    int mySqrt(int x) {
        int l{}, r{x}, ans{-1};
        while (l <= r) {
            int mid = l + (r - l) / 2;
            if ((long long)mid * mid <= x) {
                ans = mid;
                l = mid + 1;
            } else
                r = mid - 1;
        }
        return ans;
    }
};

class Solution {
public:
    /**
     * @param x: An integer
     * @return: The sqrt of x
     */
    int sqrt(int x) {
        // write your code here
        int low = 0, high = x;
        while(low <= high) {
            int mid = low + (high - low)/2;
            long long square = (long long)mid * (long long)mid;
            long long square1 = (long long)(mid + 1) * (long long)(mid + 1);
            long long square2 = (long long)(mid - 1) * (long long)(mid - 1);
 
            if(square == x) return mid;
            // (long long)mid * (long long)mid < x < (long long)(mid + 1) * (long long)(mid + 1)
            if(x > square && x < square1) return mid;
            // (long long)(mid - 1) * (long long)(mid - 1) < x < (long long)mid * (long long)mid
            if(x < square && x > square2) return mid - 1;
            if(x > square) low = mid + 1;
            else high = mid - 1;
        }
        return -1;
    }
};