Simple and easy to understand java solution

  • 36
    public class Solution {
        public int rangeBitwiseAnd(int m, int n) {
            int diffBits = 0;
            while (m != n) {
                m >>= 1;
                n >>= 1;
            return n<<diffBits;

    It's a problem that can be reduced to find the same prefix of the numbers in this range.

  • 0

    great thought to find the max prefix of binary numbers.

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