LeetCode第69題_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.

須注意當輸入是小於0的情況
都需直接回傳-1
1 返回1
0 返回0
-1 返回-1
-2 返回-1
...
依此類推


CODE

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/* * To change this license header, choose License Headers in Project Properties. * To change this template file, choose Tools | Templates * and open the template in the editor. */ package leetcode69_sqrt; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; /** * * @author chous */ public class LeetCode69_Sqrt { /** * @param args the command line arguments */ public static void main(String[] args) throws IOException { // TODO code application logic here BufferedReader input; input = new BufferedReader(new InputStreamReader(System.in)); System.out.println("Input"); String strInput = input.readLine(); int numInput = Integer.parseInt(strInput); System.out.println(mySqrt(numInput)); } public static int mySqrt(int x){ if(x==1 || x==0) return x; for(long i=1;i<=x;i++){ if(i*i > x) return (int)(i-1); } return -1; } }