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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.
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思路：
[m,n]中的所有数做“与”运算时，结果中第i位为1的条件是[m,n]中所有第i位都为1，否则为0。
因此，只需要判断m与n的最高位是否同时为1，不同时为1的话，结果为0；
同时为1时，假设这一位是第i位，这时结果为1<<i，即2^i。
java code:
public class Solution {
public int rangeBitwiseAnd(int m, int n) {
int moveCount = 0;
while(m!=n) {
m >>= 1;
n >>= 1;
moveCount++;
}
return m <<= moveCount;
}
}LeetCode:Bitwise AND of Numbers Range
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原文地址：http://blog.csdn.net/itismelzp/article/details/51637848
