• [LeetCode][Java][JavaScript]Counting Bits


    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:
    For num = 5 you should return [0,1,1,2,1,2].

    Follow up:

    • 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 in c++ or in any other language

    https://leetcode.com/problems/counting-bits/


    计算转成二进制之后有几个1。

    动态规划,二进制每多一位就把结果数组中所有的结果都加一,再放回结果数组中。

    Java:

     1 public class Solution {
     2      public static int[] countBits(int num) {
     3         if(num == 0) return new int[]{0};
     4         int[] result = new int[num + 1];
     5         int len, count = 0;
     6         while(true){
     7             len = count + 1;
     8             for(int i = 0; i < len; i++){
     9                 count++;
    10                 result[count] = result[i] + 1;                
    11                 if(count >= num)
    12                     return result;
    13             }
    14         }
    15     }
    16 }

    Javascript:

     1 /**
     2  * @param {number} num
     3  * @return {number[]}
     4  */
     5 var countBits = function(num) {
     6     if(num === 0) return [0];
     7     var result = [0], len, count = 0;
     8     while(true){
     9         len = result.length;
    10         for(var i = 0; i < len; i++){
    11             result.push(result[i] + 1);
    12             count++;
    13             if(count >= num)
    14                 return result;
    15         }
    16     }
    17 };
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  • 原文地址:https://www.cnblogs.com/Liok3187/p/5296180.html
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