leetcode笔记 201. Bitwise AND of Numbers Range

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

The key point: reduce n by removing the rightest '1' bit until n<=m;

(1)if n>m,suppose m = yyyzzz, n = xxx100, because m is less than n, m can be equal to three cases:

(a) xxx011 (if yyy==xxx)
(b) less than xxx011 (if yyy==xxx)
(c) yyyzzz (if yyy<xxx)

for case (a), and (b), xxx011 will always be ANDed to the result, which results in xxx011 & xxx100 = uuu000(uuu == yyy&xxx == xxx);

for case (c), xxx000/xxx011 will always be ANDed to the result, which results in yyyzzz & xxx000 & xxx011 & xxx100 = uuu000 (uuu <= yyy & xxx)

=> for any case, you will notice that: rangBitWiseAnd(vvvzzz,xxx100) == uuu000 == rangBitWiseAnd(vvvzzz,xxx000), (not matter what the value of"uuu" will be, the last three digits will be all zero)

=> This is why the rightest '1' bit can be removed by : n = n & (n-1);

(2)when n==m, obviously n is the result.

(3)when n < m, suppose we reduce n from rangBitWiseAnd(yyyzzz,xxx100) to rangBitWiseAnd(yyyzzz,xxx000);

i) xxx100 >yyyzzz => xxx >= yyy;

ii) xxx000 < yyyzzz => xxx <= yyy;

=> xxx == yyy;

=> rangBitWiseAnd(yyyzzz,xxx000) == rangeBitWiseAnd(xxxzzz,xxx000);

=>result is xxx000, which is also n;

 

转自https://discuss.leetcode.com/topic/20176/2-line-solution-with-detailed-explanation  charlie.chen 的回答

通过不断削减最右的位数直到n的值小于等于m,此时就能得到想要的值

似乎是默认了返回值一定是小于等于m的一个数,所以对n进行按位减少直到小于等于m

当n>m时可以证明rangBitWiseAnd(yyyzzz,xxx100) 与 rangBitWiseAnd(yyyzzz,xxx000) 结果一定相同,从而削减n

 

 

对比自己的拙劣代码:

class Solution {
public:
    int rangeBitwiseAnd(int m, int n) {
        if(m == 0){
            return 0;
        }

        int p = 1;
        int ret = 0;
       for (int i = 2; i/2 <= n && i>0;) {
        if (m%i >= i/2 && n%i <=  i - 1 && n % i >= m % i && n-m <= i/2-1)
            ret += i / 2;
        i = i * 2;
        if (i < 0)
            if (m% 2147483648 >= 1073741824 && n% 2147483648 <= 2147483647 && n % 2147483648 >= m % 2147483648 && n - m <= 1073741824 - 1)
                ret += 1073741824;
    }
        return ret;
    }
};

 

owensy的文章中对此也有相关分析 http://www.meetqun.net/thread-8769-1-1.html

但实际上复杂度似乎并没有降低多少?甚至有提高?