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[LeetCode][Java][JavaScript]Counting Bits

时间:2016-03-19 23:00:17      阅读:275      评论:0      收藏:0      [点我收藏+]

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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:

public class Solution {
     public static int[] countBits(int num) {
        if(num == 0) return new int[]{0};
        int[] result = new int[num + 1];
        int len, count = 0;
        while(true){
        	len = count + 1;
        	for(int i = 0; i < len; i++){
        		count++;
        		result[count] = result[i] + 1;        		
        		if(count >= num)
        			return result;
        	}
        }
    }
}

  

Javascript,和Java一样的代码,强行MLE,等几天看看是不是bug能否被修复:

/**
 * @param {number} num
 * @return {number[]}
 */
var countBits = function(num) {
	if(num === 0) return [0];
    var result = [0], len, count = 0;
    while(true){
    	len = result.length;
    	for(var i = 0; i < len; i++){
    		result.push(result[i] + 1);
    		count++;
    		if(count >= num)
    			return result;
    	}
    }
};

  

 

[LeetCode][Java][JavaScript]Counting Bits

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原文地址:http://www.cnblogs.com/Liok3187/p/5296180.html

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