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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.
public class Solution { public int rangeBitwiseAnd(int m, int n) { if (m>n || m==0) return 0; if (m==n) return m; int step =0; while (m!=n){ m >>= 1; n >>= 1; step++; } return m<<=step; } }
Solution 2:
If we do not know how to perform bit operator, we can still solve it like this:
public class Solution { public int rangeBitwiseAnd(int m, int n) { if (m>n || m==0) return 0; if (m==n) return m; StringBuilder mStr = new StringBuilder().append(Integer.toBinaryString(m)); StringBuilder base = new StringBuilder(); StringBuilder res = new StringBuilder(); for (int i=0;i<mStr.length();i++){ base.append(‘1‘); res.append(‘0‘); } for (int i=0;i<mStr.length();i++){ if (mStr.charAt(i)==‘0‘){ base.setCharAt(i,‘0‘); } if (mStr.charAt(i)==‘1‘){ int baseVal = Integer.parseInt(base.toString(),2); if (n>baseVal){ return Integer.parseInt(res.toString(),2); } else { res.setCharAt(i,‘1‘); } } } return Integer.parseInt(res.toString(),2); } }
LeetCode-Bitwise AND of Numbers Range
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原文地址：http://www.cnblogs.com/lishiblog/p/5867054.html
