java-判断一个自然数是否是某个数的平方。当然不能使用开方运算 - 代码物语-Let code talk - ITeye技术网站


java-判断一个自然数是否是某个数的平方。当然不能使用开方运算 - 代码物语-Let code talk - ITeye技术网站

  1.      * 题目:判断一个自然数是否是某个数的平方。当然不能使用开方运算 
  2.      * 方法1.squareRoot0  二分查找 
  3.      * 方法2.squareRoot1   
  4.      * 考虑等差数列 1 3 5 7 9...发现 
  5.      * 1^2=1 
  6.      * 2^2=1+3 
  7.      * 3^2=1+3+5 
  8.      * ... 
  9.      * 因此,N-1-3-5...若刚好可减至0,则N是某正整数的平方 
  1.     //return sqrt(n).Use binary search  
  2.     public static int squareRoot0(int n){  
  3.         int low=1;  
  4.         int high=n;  
  5.         while(low<=high){  
  6.             int mid=(low&high)+(high^low)/2;  
  7.             if(mid*mid==n){//mid*mid,overflow? I don't know how to avoid this.  
  8.                 return mid;  
  9.             }else if(mid*mid<n){  
  10.                 low=mid+1;  
  11.             }else{  
  12.                 high=mid-1;  
  13.             }  
  14.         }  
  15.         return -1;  
  16.     }  
  17.       
  18.     //return sqrt(n).  
  19.     public static int squareRoot1(int n){  
  20.         int d=1;//d=1,3,5...  
  21.         int count=0;  
  22.         while(n>0){  
  23.             n-=d;  
  24.             d+=2;  
  25.             count++;  
  26.             if(n==0){  
  27.                 return count;  
  28.             }  
  29.         }  
  30.         return -1;  
  31.     }  
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