Sqrt(x)

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69.Sqrt(x)

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.

Example 1:

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Input: x = 4
Output: 2
Explanation: The square root of 4 is 2, so we return 2.

Example 2:

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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.

个人解答：

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int mySqrt(int x) {
        long long i = 0;
        while(i*i <= x)i++;
        return i-1;
    }

二分查找：

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int mySqrt(int x) {
    int l = 0, 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;
}

牛顿迭代：

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int mySqrt(int x) {
    if (x == 0) {
        return 0;
    }

    double C = x, x0 = x;
    while (true) {
        double xi = 0.5 * (x0 + C / x0);
        if (fabs(x0 - xi) < 1e-7) {
            break;
        }
        x0 = xi;
    }
    return int(x0);
}