338. Counting Bits

　　从今天开始，每天刷一道leetcode。

　　今天的题目很简单：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.

　　初步思路：双循环，外层循环数字，内层统计‘1’出现次数。

　　新知识：ArrayList转int数组方式；

　　　　　　进制转换。

　　解决：

package com.wang.test;import java.util.ArrayList;import java.util.List;/* @Tittle * 338. 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. * @auther:wanglin * @time:2016/04/11 */public class CountingBits {  public int[] countBits(int num) {    List list = new ArrayList();    for (int n = 0; n <= num; ++n) {      // binaryNum接收转为二进制的数字      String binaryNum = Integer.toBinaryString(n);      char[] charArr = binaryNum.toCharArray();      int i = 0;      for (int m = 0; m < charArr.length; ++m) {        if (charArr[m] == '1') {          i++;        }      }      list.add(i);    }    int[] res = new int[list.size()];    for (int i = 0; i < list.size(); i++) {      res[i] = (Integer) list.get(i);    }    return res;  }  public static void main(String[] args) {    int[] res = new CountingBits().countBits(5);    for (int i : res) {      System.out.print(i);    }  }}

原标题：338. Counting Bits

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