Solution of Bitwise AND of Numbers Range

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于 2020-04-29 11:44:27 发布

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分类专栏： Leetcode 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/qq_43337384/article/details/105836297

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本文介绍了一种利用位运算解决区间[m,n]内所有数按位与的问题，通过分析区间内数字的二进制特性，提出了一种高效算法。当m和n的某一位不同时，该位及之后的所有位结果必为0。通过右移找到相同部分，再左移还原，实现O(n)时间复杂度和O(1)空间复杂度的解决方案。

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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.

Example 1:

Input: [5,7]
Output: 4

Example 2:

Input: [0,1]
Output: 0

Ideas of solving a problem
Considering the range [m, n], if n is higher than m binary digits, in the process of accumulating bitwise sum, every binary digit of the digit must have appeared 0, so once the case of different digits occurs, the result must be 0. In the program, when m and n are shifted to the right, if m and n are equal to each other, it means that the left side of the cumulative bitwise sum must be equal to m and n. At this point, we can move back to the left. The time complexity is O(n) and the space complexity is O(1). It consumes 6.1Mb of memory.

Code

package main

import (
	"fmt"
)
func main(){
	m := 5
	n := 7
	fmt.Println(rangeBitwiseAnd(m,n))
}
func rangeBitwiseAnd(m int, n int) int {
	t := 0
	for ; m != n; t++ {
		m >>= 1
		n >>= 1
	}
	return n << t
}

链接：https://leetcode-cn.com/problems/bitwise-and-of-numbers-range/solution/yi-wei-suan-fa-by-tuotuoli/
来源：力扣（LeetCode）