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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. For example, given the range [5, 7], you should return 4. Credits:Special thanks to @amrsaqr for a…
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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…
[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. 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. 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. 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,…
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…
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…
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…
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. 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…
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…
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…
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…
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…
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…
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;…
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…
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