[LeetCode] 69. Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

求平方根。

比较直观的思路就是线性扫描，但是这道题的考点应该是二分法。

二分法的具体做法是用X/2试试看是不是X的平方根，然后用two pointer上下看到底能不能找到这个平方根。如果找不到（比如26就没有平方根），就输出最接近这个数平方根的那个整数。

影子题367，思路一模一样。

时间O(log n), worse case O(n)

空间O(1)

Java实现

class Solution {
    public int mySqrt(int x) {
        // corner case
        if (x == 0) {
            return 0;
        }

        // normal case
        int low = 1;
        int high = x;
        while (low <= high) {
            long mid = low + (high - low) / 2;
            if (mid * mid == x) {
                return (int) mid;
            } else if (mid * mid < x) {
                low = (int) mid + 1;
            } else {
                high = (int) mid - 1;
            }
        }
        if (high * high < x) {
            return (int) high;
        } else {
            return (int) low;
        }
    }
}

JavaScript实现

/**
 * @param {number} x
 * @return {number}
 */
var mySqrt = function(x) {
    // corner case
    if (x <= 0) {
        return 0;
    }

    // normal case
    let low = 1;
    let high = x;
    while (low <= high) {
        let mid = parseInt(low + (high - low) / 2);
        if (mid * mid === x) {
            return mid;
        } else if (mid * mid < x) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    if (high * high < x) {
        return high;
    } else {
        return low;
    }
};

LeetCode 题目总结