201. Bitwise AND of Numbers Range

最新推荐文章于 2024-01-19 04:58:25 发布

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最新推荐文章于 2024-01-19 04:58:25 发布

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原文链接：http://www.cnblogs.com/neweracoding/p/5722316.html

本文提供了一种解决LeetCode上201题的方法，即求解一个给定范围[m,n]内所有数字的按位与结果。通过不断右移m和n直到它们相等，并记录下右移次数，最后将该次数作为左移位数乘以m得到最终结果。

201. 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.

Bit Manipulation

public class Solution {
  public int rangeBitwiseAnd(int m, int n) {
    if (m == 0)
      return 0;

    int powerOf2 = 1;
    while (m != n) { //if m!=n, current bit result is 0.
      m >>= 1;
      n >>= 1;
      powerOf2 <<= 1;
    }
    return m * powerOf2;
  }
}