标签：
Bitwise AND of Numbers Range
问题：
Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.
思路：
分治法
我的代码：
public class Solution { public int rangeBitwiseAnd(int m, int n) { if( m>n || m<=0 || n<=0) return 0; if(m == n) return m; if(m == n-1) return m&n; int mid = (m+n)/2; return mid & rangeBitwiseAnd(m,mid-1) & rangeBitwiseAnd(mid+1,n); } }
他人代码：
public class Solution { public int rangeBitwiseAnd(int m, int n) { if(m == 0){ return 0; } int moveFactor = 1; while(m != n){ m >>= 1; n >>= 1; moveFactor <<= 1; } return m * moveFactor; } }
学习之处：
标签：
原文地址：http://www.cnblogs.com/sunshisonghit/p/4461390.html
