leetcode 69. Sqrt(x) 367. Valid Perfect Square-优快云博客

本文链接：https://blog.youkuaiyun.com/liqinzhe11/article/details/78579742

本文介绍如何使用二分法求解整数的平方根，并判断一个数是否为完全平方数。针对两种问题提供了具体的算法实现示例。

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69. Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

x is guaranteed to be a non-negative integer.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since we want to return an integer, the decimal part will be truncated.

二分法：二分答案。

使用二分法标准格式。

class Solution {
public:
    int mySqrt(int target) 
    {
        int start = 0;
        int end = min(target, 46340);
        while (start + 1 < end)
        {
            int mid = (end - start) /2 + start;
            if (mid * mid == target)
                return mid;
            else if (mid * mid > target)
                end = mid;
            else
                start = mid;
        }
        //double check
        if (end * end <= target)
            return end;
        else
            return start;  
    }
};

367. Valid Perfect Square

Given a positive integer num, write a function which returns True if num is a perfect square else False.

Note: Do not use any built-in library function such as sqrt.

Example 1:

Input: 16
Returns: True

Example 2:

Input: 14
Returns: False

和上面的题做法一模一样。二分法。

class Solution {
public:
    bool isPerfectSquare(int num) 
    {
        if (num <= 1) return true;
        int start = 1, end = min(num, 46340);
        while (start + 1 < end)
        {
            int mid = (end - start) / 2 + start;
            if (mid * mid == num) return true;
            else if (mid * mid < num) start = mid;
            else    end = mid;
        }
        //double check
        return (start * start == num || end * end == num) ;
    }
};