public int[] countBits(int num) { int[] arr = new int[num + 1]; arr[0] = 0; if (num <= 0) { return arr; } arr[1] = 1; //当前执行到的数 int index = 1; //当前到达的平方数 int powNum = 0; //上次到达的数 int powSum = (int) Mat
P(x)=P(x&(x−1))+1; java实现的源码 publicclassSolution {publicint[] countBits(intnum) {int[] ans =newint[num + 1];for(inti = 1; i <= num; ++i) ans[i]= ans[i & (i - 1)] + 1;returnans; } } 源码来源:https://leetcode-cn.com/problems/counting-bits/solution/bi-t...
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}...
class Solution { public: vector<int> countBits(int num) { vector<int> dp(num+1,0); for(int i=1;i<=num;i<<=1){ for(int j=0;j<i;j++){ if(i+j<=num) dp[i+j] = dp[j]+1; } } return dp; } }; 1 2 3 4 5 6 7 8 9 10 11 12版权...
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 new int[0]; int[] res=new int[num+1]; Arrays.fill(res, 0); for(int i=1;i<=num;i++) ...
LeetCode "Counting Bits" Neat DP problem to go. frommathimport*classSolution(object):defcountBits(self, num): ret=[0]foriinrange(1, num + 1):ifi & 1:#oddret += [ret[-1] + 1]else: prev= i - 1trailing1s= (i | prev) -i...
LeetCode 338. Counting Bits 简介:给定一个非负整数 num。对于 0 ≤ i ≤ num 范围中的每个数字 i ,计算其二进制数中的 1 的数目并将它们作为数组返回。 Description Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in ...
It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass? Space complexity should be O(n). Can you do it like a boss? Do it without using any builtin function like __builtin_popcount ...
publicint[]countBits(intnum){int[]res=newint[num+1];for(inti=0;i<res.length;i++)res[i]=res[i>>>1]+(i&1);returnres;}
class Solution { public: vectorcountBits(int num) { //一个数组有(0~num)即num+1个元素,初始化为0 vectorv1(num+1,0); for(int i=1;i<=num;i++) { v1[i]=v1[n&(n-1)]+1; } } } 网页题目:leetcode(1)--338.CountingBits ...