int rangeBitwiseAnd(int m, int n) { a[0]=1;for(int i=1;i<=30;i++) { a[i]=a[i-1]*2; } int ans=0;while(1) { int x =fun(m,n);if(x==0)break; ans+=x; m -=x; n -=x; }returnans; } intfun(int x,int y){for(int i=30;i>=0;i--) {if(x<a[i]&&y>=a[i]) {return0; }if(x>=a...
Given two integers left and right that represent the range [left, right], return the bitwise AND of all numbers in this range, inclusive. 范围内的数字按照位数与 暴力法 超时 classSolution {public:intrangeBitwiseAnd(intleft,intright) {intres=left;for(inti=left+1;i<=right;++i){ res&=i; ...
LeetCode 201. Bitwise AND of Numbers Range 程序员木子 香港浸会大学 数据分析与人工智能硕士在读 来自专栏 · LeetCode Description 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]...
Bitwise AND of Numbers Range https://leetcode.com/problems/bitwise-and-of-numbers-range/description/ 嗯。这一题,看上去好麻烦啊。但实际上就一点:In one word, this problem is asking us to find the common prefix of m and n 's binary code. 上面那句话照抄了某个dicu...Bitwise AND of ...
Can you solve this real interview question? Bitwise AND of Numbers Range - Given two integers left and right that represent the range [left, right], return the bitwise AND of all numbers in this range, inclusive. Example 1: Input: left = 5, right =
简介:老程序员分享:leetcode笔记201.BitwiseANDofNumbersRange " public int rangeBitwiseAnd(int m, int n) { while(m //代码效果参考:https://v.youku.com/v_show/id_XNjQwMDEyOTc4MA==.html return n; } The key point: reduce n by removing the rightest '1' bit until n<=m; ...
LeetCode 201. Bitwise AND of Numbers Range(位运算) 题目 题意:给你两个数n,m 0<= n<=m <=2^31-1 ,让你计算从n到m的每个数依次位与的结果。 题解:当然不能for循环,按位运算。会超时。 经过分析我们发现当一个数字为2的整次幂,比如8 :1000,8往后位与的结果都是8,直到位与到16 ...
Runtime: 5 ms, faster than 100.00% of Java online submissions for Bitwise AND of Numbers Range. class Solution { public int rangeBitwiseAnd(int m, int n) { int result = m & n; for (int i = 0; m != n; i++, m >>= 1, n >>= 1) { ...
Leetcode 201. Bitwise AND of Numbers Range Bitwise AND of Numbers Range 题目大意:给定n,m两个数,要求你对n到m这个区间的所有数做且运算,问最后的结果是多少(不得不说,最近这些天的每日打卡题目质量都还挺好的) 题目思路:如果对区间所有数进行且运算,显然不合理,我们考虑这样一种情况,因为是做...
publicintrangeBitwiseAnd(intm,intn){returnm==n?m:m&-Integer.highestOneBit(m^n);} 我们调用了库函数Integer.highestOneBit,我们去看一下它的实现。 /*** Returns an {@code int} value with at most a single one-bit, in the* position of the highest-order ("leftmost") one-bit in the specif...