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  • LeetCode 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.

     1 class Solution {
     2 public:
     3     int rangeBitwiseAnd(int m, int n) {
     4         int res = 0;
     5         
     6         for (int i=0; i<32 && (n!= 0 && m != 0); i++) {
     7             res = res | (bitvalue(m, n) << i);
     8             m>>=1;
     9             n>>=1;
    10         }
    11         return res;
    12     }
    13     
    14     int bitvalue(int m, int n) {
    15         if (n - m > 0) {
    16             return 0;
    17         }
    18         if (n == m && (n&0x1 == 1)) {
    19             return 1;
    20         }
    21         return 0;
    22     }
    23 };

    都忘了自己第一次是怎么想出来的了,看了discuss的一种解法:

    class Solution {
    public:
        int rangeBitwiseAnd(int m, int n) {
            int x = m^n;
            int cnt = 0;
            while (x) {
                cnt++;
                x>>=1;
            }
            return (m & n) & ~((1<<cnt) - 1);
        }
    };

    简单很多

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  • 原文地址:https://www.cnblogs.com/lailailai/p/4430972.html
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