LeetCode 338 Counting Bits 原题链接: Counting Bits 题目大意: 输入一个数字num,输出一个数组,表示从0到num的所有数字的二进制形式中,1的个数。 解题思路: 找规律。 0: [0] 1: [0,1+a[0]] 2: [0,1+a[0],1+a[0]] 3: [0,1+a[0],1+a[0],1+a[1]] 4: [0,1+a[0],1+a[0...
【LeetCode】338. Counting Bits (2 solutions) Counting Bits 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: Fornum = 5you should return[0,1,1,2,...
https://leetcode.com/problems/counting-bits/ https://leetcode.com/discuss/92796/four-lines-c-time-o-n-space-o-1 https://leetcode.com/discuss/92694/my-408-ms-c-solution-using-bitset https://leetcode.com/discuss/92698/my-448ms-c-easy-solution-o-n-time-and-o-n-space LeetCode All i...
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Medium, Acceptance为19%。 题目如下: Given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method to set all bits between i and jin N equal ......
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...
Counting Bits https://leetcode.com/problems/counting-bits/ 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....
//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...
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; } };
338. Counting Bits Difficulty: Medium 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 r...Leetcode 338. Counting Bits 题目 链接:https://leetcode.com/problems/counting-bits Level: ...
[leetcode]338. Counting Bits solution 1: 因为乘以2相当于左移,最右添0,所以1的个数都不变。 (这一句的思想是精髓) 奇数i的个数相当于是前一个偶数,末尾+1得到的。而前一个偶数中1的个数=这个偶数/2的1的个数。 又因为偶数和其后一个奇数除以二,都得到同一个偶数。(比如4和2的个数一样,5/2...