201. Bitwise AND of Numbers Range

原创于 2017-09-08 11:14:40 发布 · 184 阅读

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本文介绍了一种利用位运算解决特定数学问题的方法——求解一个闭区间内所有整数的按位与结果。通过逐步右移区间两端的数值直至它们相等，再将结果左移相同次数，即可高效获得答案。

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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.


last bit of (odd number & even number) is 0.
when m != n, There is at least an odd number and an even number, so the last bit position result is 0.
Move m and n rigth a position.
找到m,n左边公共的部分即可
*/

class Solution {
public:
    int rangeBitwiseAnd(int m, int n) {
        int i=0;
        while(m != n)
        {
            m>>=1;
            n>>=1;
            i++;
        }
        return m<<i;
    }
};