https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/56729/Bit-operation-solution(JAVA) 面试官,你再问我 Bit Operation 试试? 描述 Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusi…
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).例如,给定范围 [5,7],您应该返回 4. 详见:https://leetcode.com/problems/bitwise-and-of-numbers-range/description/ Java实现: 将m和n中的所有整数相与,得到的结果是m和n中的所有数的共同高位保留,除共同高位之外的其他位置零.那么关键问题就是如何找到这个共同的高位,其实并不难…
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 因为i和i + 1只相差1,所以两个数二进制的最后一位一定不一样,最后一位按位与结果一定是0.所以n,m按位与的最后一位肯定是0,我们只需要计算除去最后一位的二进制按位与的结果.即除以2后按位与的结果. class Solution { public: int r…
[LeetCode]201. Bitwise AND of Numbers Range 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/bitwise-and-of-numbers-range/description/ 题目描述: Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers i…
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. Credits:Special thanks to @amrsaqr for a…
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. Example 1: Input: [5,7] Output: 4 Example 2: Input: [0,1] Output: 0 解法1 一个一个做位与,但是会超时... 解法2 事实: 与运…
在Java位运算总结-leetcode题目博文中总结了Java提供的按位运算操作符,今天又碰到LeetCode中一道按位操作的题目 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. 题意:…
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 creatin…
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. 解题思路: 本题有很多思路,最简单的方法: result就是m和n二进制前面相同的部分!!! JAVA实现如下: public int ra…
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. int rangeBitwiseAnd(int m, int n) { int mask = 0xffffffff; /* find out…
题目: 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. 链接: http://leetcode.com/problemset/algorithms/ 题解: 一开始采用暴力解,自然超时了.…
Problem: 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. Analysis: The idea behind this problem is not hard, you could…
Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND(按位与) of all numbers in this range, inclusive. For example, given the range [5, 7], you should return 4. 思路:序列是按1递增的,所以必定先影响低位,按位与为1的情况必定发生在高位.两个数向右移,当两数相等,剩余的1便是按位与得到的1.…
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. 题解:如果m==n,那么答案就是m. 如果m<n,那么二进制最右边一位在最后的结果中肯定是0,那么就可以转化成子问题: rangeBi…
https://leetcode.com/problems/bitwise-and-of-numbers-range/ [n,m]区间的合取总值就是n,m对齐后前面一段相同的数位的值 比如 5:101 7:111 结果就是 4:100 class Solution { public: int rangeBitwiseAnd(int m, int n) { int len = 0; long long tmp = 1; for(;tmp <= n || tmp <= m;len++){tmp&l…
201. 数字范围按位与 201. Bitwise AND of Numbers Range 题目描述 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). LeetCode201. Bitwise AND of Numbers Range中等 示例 1: 输入: [5,7] 输出: 4 示例 2: 输入: [0,1] 输出: 0 Java 实现 方法一 class Solution { public…
一:Number of 1 Bits 题目: Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight). For example, the 32-bit integer '11' has binary representation 00000000000000000000000000001011, so t…
1. Number of Islands Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid a…
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. Credits: Special thanks to @amrsaqr for a…
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…
题目 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. 分析 题目字面意思是给定两个整数构成闭区间, 0 <= m <= n <= 2147483647 均为合法整数,求区…
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. 题意: 给定 [m, n] 范围内个数,返回范围内所有数相与的结果. 思路: 如果打着暴力的旗号,那么这个题是会超时的 - -!.最后也是没…
2018-08-13 22:50:51 问题描述: 问题求解: 首先如果m 和 n不相等,那么必然会有至少一对奇偶数,那么必然末尾是0. 之后需要将m 和 n将右移一位,直到m 和 n相等. 本质上,本题就是求m 和 n的最长preSubNum. public int rangeBitwiseAnd(int m, int n) { if (m == 0) return 0; int moveFactor = 1; while (m != n) { m >>= 1; n >>= 1;…
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. 分析:http://www.cnblogs.com/grandyang/p/4431646.html 我们先从题目中给的例子来分析,[5,…
题目: 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. 解答: 假如说5:101  7:111  连续几个数相与的规律:一,仅仅要是同样的位置的数字不同样最后那个位置的结果一定是0 .二,…
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: 先找前面二进制相同的位,后面不相同的位相与一定为0. 比如: 1111001 1110011 从0011 - 1001 中间一定会经…
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. 题目大意:给一个范围,返回这个范围中所有的数按位相与最后的结果. 解题思路:当拿到这个题目的时候,我是拒绝的,这么简单的题,直接与然后返回不…
C#版 - Leetcode 201. 数字范围按位与(bitwise AND) - 题解 在线提交: https://leetcode.com/problems/bitwise-and-of-numbers-range/ 题目描述 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 题目难度:Medium 通过次…
201. 数字范围按位与 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 class Solution { public int rangeBitwiseAnd(int m, int n) { while (m < n) n &= n - 1; return n; } }…
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 思路分析 由于是按位与,那么某位一旦出现0,结果该位肯定是0.所以只需要考虑m,n都是1的位置.那么直接从高位开始,往低位走,直到遇到该为的数字不相等,将其后的数为都置为0,即为[m,n]之间所有的数字按位与的结果.代码如下 #include<bits/st…