201. Bitwise AND of Numbers Range

Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.Example 1:Input: [5,7]Output: 4Example 2:Input: [0,1]Output: 0难度：medium题目：给定范围 [m, n] 0 <= m <= n <= 2147483647, 返回这个范围内所有数的与计算结果。思路：m 到 n 之间所有数的位与计算，可归结为找出最左边的不同的位即（1, 0) （0, 1)，然后把当前位与之后的位都置成0.例如，二进制数ABCD EFGH IGKL MN1001 0001 1000 01,1010 0110 1010 00最左边的不同位为C位，C位及C位之后的位都必然经过0,1 之间的变换，因此位与计算就等于将C位及之后的所有位置成0.Runtime: 5 ms, faster than 100.00% of Java online submissions for Bitwise AND of Numbers Range.class Solution { public int rangeBitwiseAnd(int m, int n) { int result = m & n; for (int i = 0; m != n; i++, m >>= 1, n >>= 1) { if (1 == ((m ^ n) & 1)) { result = result >> i << i; } } return result; }}