[leetcode] 201. Bitwise AND of Numbers Range解题报告

题目链接: https://leetcode.com/problems/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.

For example, given the range [5, 7], you should return 4.

思路: 有两种思路来做这题, 第一种是从后往前每次 n = (n & (n-1)), 直到n <= m, 然后返回n, 这种方法的思路是可以逐渐将最右边不同的置为0, 最后只剩下最左边m 和 n相同的位置.

还有一种方法是先求出n-m, 我们可以根据n-m求出m和n最右边不相同的有多少位, 即log(n-m)/ log2+1 位也就是剩余的左边的位是相同的, 所以我们只要将其左边相同的位数求出来即可.

两种代码如下:

1.

class Solution {
public:
    int rangeBitwiseAnd(int m, int n) {
        while(n > m) n &= (n-1);
        return n;
    }
};

参考: https://leetcode.com/discuss/91952/my-c-solution-with-explanation

2.

class Solution {
public:
    int rangeBitwiseAnd(int m, int n) {
        int cnt = log(n-m)/log(2) + 1;
        int k = INT_MAX ^ ((1<<cnt)-1);
        return k & m & n;
    }
};

参考: https://leetcode.com/discuss/91966/my-c-solution-2-with-explanation