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  • [leetcode] Bitwise AND of Numbers Range

    Bitwise AND of Numbers Range

    Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.

    For example, given the range [5, 7], you should return 4.

    Credits:
    Special thanks to @amrsaqr for adding this problem and creating all test cases.

    Have you met this question in a real interview?
     
    思路:
    找规律即可,通过分析我们发现:
    1,当 m = n时,直接返回m或者n即可。
    2,当表示m和n的二进制的数的长度不相等时,由于低一位像高一位进位时,出现的数必然为100...0,长度比低位的数长了一位,此数与长度比它小的数“与”,结果必然为0,0“与”任何数都为0。即,当表示m,n的二进制数的长度不一致时,从m到n进行“与”操作,结果为0;
    3,当二者长度一致时,进行循环与操作即可。只是要注意:n为INT_MAX 时,需要特殊处理。
     
    class Solution
    {
    public:
      int rangeBitwiseAnd(int m, int n)
      {
        int ret = m;
        int a = m, b = n;
        int lena = 1, lenb = 1;
    
        if(m == n)
          return m;
    
        while(a >>= 1)
          lena++;
    
        while(b >>= 1)
          lenb++;
    
        if(lena == lenb)
        {
          for(int i = m + 1; i <= n; i++)
          {
            ret &= i;
            if(i == INT_MAX)
              break;
          }
        }
        else
          ret = 0;
         
        return ret; 
      }
    };
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  • 原文地址:https://www.cnblogs.com/lxd2502/p/4479970.html
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