LeetCode-69：Sqrt(x) (整數的平方根)

題目：

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.

例子

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.

問題解析：

求給定整數x的平方根。

鏈接：

• https://leetcode.com/problems/sqrtx/description/

思路標籤：

算法：二分法

解答：

1. 二分法

• 使用二分法在1~x的範圍內進行查找；
• 雖然題目簡單，但是需要注意的邊界範圍，以及mid的賦值方式需要考慮；

class Solution {
public:
    int mySqrt(int x) {
        if (0 == x) return 0;
        int left = 1, right = x, ans;
        while (left <= right) {
            int mid = left + (right - left) / 2; //mid爲在原來的基礎上+一半
            if (mid <= x / mid) {
                left = mid + 1;
                ans = mid;
            } else {
                right = mid - 1;
            }
        }
        return ans;
    }
};

2. 牛頓法

• 直接利用牛頓法求解。
• 具體解釋自行Google。

class Solution {
public:
    int mySqrt(int x) {
        long r = x;
        while (r*r > x)
            r = (r + x/r) / 2;
        return r;
    }
};