LeetCode //C - 69. Sqrt(x)

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.

For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python.

Example 1:

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

Example 2:

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.

Constraints:

• 0 < = x < = 2 31 − 1 0 <= x <= 2^{31} - 1 0<=x<=231−1

From: LeetCode

Link: 69. Sqrt(x)

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Solution:

Ideas：

• We first handle the special case of x being 0.
• We define left and right to represent the range of possible square root values.
• Inside the while loop, we calculate the mid of the range.
• We compare mid * mid with x. To avoid integer overflow, we use mid <= x / mid.
• If mid * mid is less than or equal to x, we update left and ans.
• If mid * mid is greater than x, we adjust right.
• When the loop finishes, ans contains the floor of the square root of x.

Code:

int mySqrt(int x) {
    if (x == 0) return 0;
    int left = 1, right = x, ans;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (mid <= x / mid) { // To avoid overflow
            left = mid + 1;
            ans = mid;
        } else {
            right = mid - 1;
        }
    }
    return ans;
}