Do it without using any builtin function like __builtin_popcount in c++ or in any other language. 题意很简单,直接位运算即可。 代码如下: import java.util.Arrays; /* * 第一种方法感觉想不出来,但是看起来很对 * */ class Solution { public int[] countBits(int num) { if(num<0) return ...
然后发现代码又长,效率不怎样,这里splice比起在vector中计算元素的位置没有什么优势,问题在于每次push_back都要新申请一个元素的空间,最后还要重新转换成vector(PS:leetcode给出的C++函数签名的返回值是vector),这种靠链接list的小伎俩还不如老老实实用vector。(好吧我承认是被STL的list排序算法影响到了……) 于是...
LeetCode: Counting Bits f[i] = f[i/2] + i%2; 1publicclassSolution {2publicint[] countBits(intnum) {3int[] ans =newint[num+1];4for(inti = 0; i < ans.length; i++) {5ans[i] = ans[i / 2] + i % 2;6}7returnans;8}9}...
LeetCode 338. Counting Bits 今天开始,刷leetcode!一方面重新熟练编程,一方面为面试做准备! 同时,记录每道题,并且要解释清楚,为的是面试能解释给面试官听! 题意:求出【0,1,2,3……num】他们每个数字的二进制的1的个数 思路: 用举例法 8的二进制是1000, 15的二进制是1111 即8~15的二进制是1000~1111 ...
//c++版本 class Solution { public: vectorcountBits(int num) { //一个数组有(0~num)即num+1个元素,初始化为0 vector v1(num+1,0); for(int i=1;i<=num;i++) { v1[i]=v1[n&(n-1)]+1; } } } 本文名称:leetcode(1)--338.CountingBits 文章位置:http://www.cdysf.com/article...
Counting Bits 题目Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array. 答案...Update Bits Update Bits 今天的题目来自LintCode,是一道关于数学和位运算的题目,难度为...
Counting Bits 题目内容 Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as ...LeetCode --- 338. Counting Bits 338. Counting Bits Difficulty: Medium Given a non ...
【LeetCode】338. Counting Bits like __builtin_popcountinc++ orinany other language. 题目:给定一个非负整数n,计算[0,n]区间每个数二进制中1bit的个数。 要求:时间...想到,但没肯定,当时只想着怎么去除最高位1。。。 代码:classSolution{ //338.CountingBitspublicint[]countBits(intnum ...
ret[i] = ret[i & (i - 1)] + 1. Code classSolution{public:vector<int>countBits(intnum){vector<int>table(num +1,0);for(inti =1; i <= num; i++) { table[i] = table[i & (i -1)] +1; }returntable; } };
Java [Leetcode 338]Counting Bits 题目描述: Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array. Example:For num = 5 you should return [0,1,1,2,1,2]. ...