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.
解題思路：基數(shù)和偶數(shù)的and結(jié)果末位為0。為了找m，n之間的公共前綴，mn分別右移，并記錄次數(shù)。
public class Solution { public int rangeBitwiseAnd(int m, int n) { if (m == 0) return 0; int factor = 1; while (m != n) { m >>= 1; n >>= 1; factor <<= 1; } return m*factor; }}public class Solution { public int rangeBitwiseAnd(int m, int n) { if (m == 0) return 0; int step = 0; while (m != n) { m >>= 1; n >>= 1; step++; } return (m<<step); }}public class Solution { public int rangeBitwiseAnd(int m, int n) { if (m == 0) return 0; int step = 0; while (m < n) { n &= (n-1); } return n; }}新聞熱點
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