C++实现LeetCode(201.数字范围位相与)

[LeetCode] 201.Bitwise AND of Numbers Range 数字范围位相与

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.

Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.

又是一道考察位操作Bit Operation的题,相似的题目在LeetCode中还真不少,比如Repeated DNA Sequences 求重复的DNA序列, Single Number 单独的数字,   Single Number II 单独的数字之二 , Grey Code 格雷码,和 Reverse Bits 翻转位 等等,那么这道题其实并不难,我们先从题目中给的例子来分析,[5, 7]里共有三个数字,分别写出它们的二进制为:

101  110  111

相与后的结果为100,仔细观察我们可以得出,最后的数是该数字范围内所有的数的左边共同的部分,如果上面那个例子不太明显,我们再来看一个范围[26, 30],它们的二进制如下:

11010  11011  11100  11101  11110

发现了规律后,我们只要写代码找到左边公共的部分即可,我们可以从建立一个32位都是1的mask,然后每次向左移一位,比较m和n是否相同,不同再继续左移一位,直至相同,然后把m和mask相与就是最终结果,代码如下:

解法一:

class Solution {
public:
    int rangeBitwiseAnd(int m, int n) {
        int d = INT_MAX;
        while ((m & d) != (n & d)) {
            d <<= 1;
        }
        return m & d;
    }
};

此题还有另一种解法,不需要用mask,直接平移m和n,每次向右移一位,直到m和n相等,记录下所有平移的次数i,然后再把m左移i位即为最终结果,代码如下:

解法二:

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

下面这种方法有点叼,就一行搞定了,通过递归来做的,如果n大于m,那么就对m和n分别除以2,并且调用递归函数,对结果再乘以2,一定要乘回来,不然就不对了,就举一个最简单的例子,m = 10, n = 11,注意这里是二进制表示的,然后各自除以2,都变成了1,调用递归返回1,这时候要乘以2,才能变回10,参见代码如下:

解法三:

class Solution {
public:
    int rangeBitwiseAnd(int m, int n) {
        return (n > m) ? (rangeBitwiseAnd(m / 2, n / 2) << 1) : m;
    }
};

下面这种方法也不错,也很简单,如果m小于n就进行循环,n与上n-1,那么为什么要这样与呢,举个简单的例子呗,110与上(110-1),得到100,这不就相当于去掉最低位的1么,n就这样每次去掉最低位的1,如果小于等于m了,返回此时的n即可,参见代码如下:

解法四:

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

参考资料:

https://discuss.leetcode.com/topic/13508/one-line-c-solution

https://discuss.leetcode.com/topic/12133/bit-operation-solution-java

https://discuss.leetcode.com/topic/20176/2-line-solution-with-detailed-explanation

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