[LeetCode] Sqrt(x)

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最新推荐文章于 2024-05-28 17:22:45 发布 · 112 阅读

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本文提供了一种计算整数平方根的有效算法实现。通过使用二分查找法，该算法能够在较短的时间内找到最接近目标值的平方根。代码中展示了两种不同的实现方式，并考虑了边界条件。

Implement int sqrt(int x).

Compute and return the square root of x.

class Solution {
public:
    int sqrt(int x) {
        int l = 0, r = 46340;
        if (r*r <= x) return r;
        while (l < r) {
            if (r - l < 2) {
                if (r*r == x) return r;
                return l;
            }
            int m = (r+l) / 2;
            int t= m*m;
            if (t == x) return m;
            else if (t > x) r = m;
            else l = m;
        }
    }
};

@2013-10-07

感觉有点多此一举。

class Solution {
public:
    int sqrt(int x) {
        int left = 0, right = 1;
        while (right * right < x) {
            if (right * right > INT_MAX / 4) {
                left = right;
                // sqrt(INT_MAX) = 46340.9
                right = 46340;
                break;
            }
            left = right;
            right *= 2;
        }
        int mid, t;
        while (right - left > 1) {
            mid = (right + left) / 2;
            t = mid * mid;
            if (t == x) return mid;
            else if (t < x) left = mid;
            else right = mid;
        }
        if (right * right > x) return left;
        else return right;
    }
};