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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0069_sqrtx;

// #Easy #Top_Interview_Questions #Math #Binary_Search #Binary_Search_I_Day_4
// #2023_08_11_Time_1_ms_(99.51%)_Space_39.5_MB_(78.13%)

/**
 * 69 - Sqrt(x)\.
 *
 * Easy
 *
 * Given a non-negative integer `x`, compute and return _the square root of_ `x`.
 *
 * Since the return type is an integer, the decimal digits are **truncated** , and only **the integer part** of the result is returned.
 *
 * **Note:** You are not allowed to use any built-in exponent function or operator, such as `pow(x, 0.5)` or `x ** 0.5`.
 *
 * **Example 1:**
 *
 * **Input:** x = 4
 *
 * **Output:** 2 
 *
 * **Example 2:**
 *
 * **Input:** x = 8
 *
 * **Output:** 2
 *
 * **Explanation:** The square root of 8 is 2.82842..., and since the decimal part is truncated, 2 is returned.
 *
 * **Constraints:**
 *
 * *   0 <= x <= 2³¹ - 1
**/
public class Solution {
    public int mySqrt(int x) {
        int start = 1;
        int end = x / 2;
        int sqrt = start + (end - start) / 2;
        if (x == 0) {
            return 0;
        }
        while (start <= end) {
            if (sqrt == x / sqrt) {
                return sqrt;
            } else if (sqrt > x / sqrt) {
                end = sqrt - 1;
            } else if (sqrt < x / sqrt) {
                start = sqrt + 1;
            }
            sqrt = start + (end - start) / 2;
        }
        if (sqrt > x / sqrt) {
            return sqrt - 1;
        } else {
            return sqrt;
        }
    }
}