[Swift]LeetCode201. 数字范围按位与 | Bitwise AND of Numbers Range

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10184197.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.

Example 1:

Input: [5,7]
Output: 4

Example 2:

Input: [0,1]
Output: 0

---

给定范围 [m, n]，其中 0 <= m <= n <= 2147483647，返回此范围内所有数字的按位与（包含 m, n 两端点）。

示例 1: 

输入: [5,7]
输出: 4

示例 2:

输入: [0,1]
输出: 0

---

44ms

class Solution {
    func topDigit(_ n: Int) -> Int {
        var n = n
        while (n & (n-1)) != 0 {
            n = (n & (n-1))
        }
        return n
    }
    func rangeBitwiseAnd(_ m: Int, _ n: Int) -> Int {
        guard m > 0 && n > 0 else {
            return 0
        }
        var m2 = topDigit(m)
        var n2 = topDigit(n)
        
        if n2 != m2 {
            return 0
        }
        
        return m2 | rangeBitwiseAnd(m-m2, n-m2)
    }
}

---

48ms

class Solution {
    func rangeBitwiseAnd(_ m: Int, _ n: Int) -> Int {
        if m == 0 {
            return 0
        }
        
        var m = m
        var n = n
        var factor = 1
        
        while m != n {
            m >>= 1
            n >>= 1
            factor <<= 1
        }
        
        return m * factor
    }
}

---

52ms

class Solution {
    func rangeBitwiseAnd(_ m: Int, _ n: Int) -> Int {
        /*
        0  - 0000
        1  - 0001
        2  - 0010
        3  - 0011
        4  - 0100
        5  - 0101
        6  - 0110
        7  - 0111
        8  - 1000
        9  - 1001
        10 - 1010
        
        8..10  =>  1000
        */

        var m = m
        var n = n
        var i = 0
        while m != n {
            m >>= 1
            n >>= 1
            i += 1
        }
        return m << i
    }
}

---

52ms

class Solution {
    func rangeBitwiseAnd(_ m: Int, _ n: Int) -> Int {
        return (n > m) ? (rangeBitwiseAnd(m / 2, n / 2) << 1) : m

    }
}

---

56ms

class Solution {
    func rangeBitwiseAnd(_ m: Int, _ n: Int) -> Int {
        var plus = 1
        var result = m & n
        while m + plus < n {
            result &= m + plus
            plus += plus
        }
        return result
    }
}

---

64ms

class Solution {
    func rangeBitwiseAnd(_ m: Int, _ n: Int) -> Int {
        // 101011101
        // 101111001
        
        var m = m
        var n = n
        var res = 0
        var t = 1
        for i in 0..<31 {
            if m % 2 == 0 || n % 2 == 0 || m < n {
                m = m >> 1
                n = n >> 1
            } else {
                res += t
                m = m >> 1
                n = n >> 1
            }
            
            t *= 2
        }
        
        return res
    }
}