【Leetcode】Bitwise AND of Numbers Range

题目链接：https://leetcode.com/problems/bitwise-and-of-numbers-range/

题目：

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.

思路：

32位整型，只需判断每一位在m~n之间是否有0存在，若有，该位 在AND操作中一定为0，只需判断32次

算法：

 public int rangeBitwiseAnd(int m, int n) {
         if (m == n)
             return m;
         int res = 0, i = 0;
         while (n != 0 && m != 0) {
             // 检查m~n是否有偶数存在时间复杂度O(1)
             boolean flag = false;
             for (int j = m; j <= n; j++) {
                 int tmp = j;
                 if ((tmp & 1) == 0) {// 最后一位为0
                     flag = true;
                     break;
                 }
             }

             if (flag == false) {// 最低位全是1
                 res = res | (1 << i); // 将指定为置为1
             }
             i++;
             n = n >> 1;
             m = m >> 1;
         }
         return res;
     }