Leetcode 69, @ Java－CONY的世界

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.

這題我只想到暴力解，所以就直接放棄看解答了，解法非常有趣

用BFS，不斷逼近正確的區段，因為有可能不是整數，最後確定數值是用 if(mid+1 > x/(mid+1)) 來判斷（來確定下個值會過大，正解就是這個值）

class Solution {
    public int mySqrt(int x) {
        int l=1, r=(int)Math.ceil(x/2.0);
        while(l <= r) {
            int mid = l+(r-l)/2;
            if(mid > x/mid)
                r = mid-1;
            else {
                if((mid+1) > x/(mid+1))
                    return mid;
                l=mid+1;
            }
        }
        return 0;
    }
}