Leetcode 69. Sqrt(x)耍流氓的代码呀

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.

题目链接：https://leetcode.com/problems/sqrtx/

自己写了一份代码，然后看别人，这tm代码都能过，不超时？？？？耍流氓啊

class Solution {
public:
    int mySqrt(int x) {
        long start=0,end=x/2,mid=0;
        while(start<=end)
        {
            mid=(start+end)/2;
            if(mid*mid==x)
            {
                return mid;
            }
            else if(mid*mid<x)
            {
                start=mid+1;
            }
            else
            {
                end=mid-1;
            }
        }
        return start*start>x?start-1:start;
    }
};

 

看看人家耍流氓的代码！！！！

class Solution {
public:
    int mySqrt(int x) {
        int i=0;
        for(;i<=46340;++i)
            if(i*i>x)return i-1;
            else if(i*i==x)return i;
        return i-1;
    }
};

这都能过，出这题还有意思吗？？？？？