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Leet Code OJ 338. Counting Bits [Difficulty: Easy]

时间:2016-03-18 21:56:44      阅读:208      评论:0      收藏:0      [点我收藏+]

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题目:
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.
Hint:
You should make use of what you have produced already.

翻译:
给定一个非负整数num,对于每个0<=i<=num的整数i,计算i的二进制表示中1的个数,返回这些个数作为一个数组。
例如,输入num = 5 你应该返回 [0,1,1,2,1,2].

分析:
按照常规思路,很容易得出“Java代码2”的方案,但是这个方案的时间复杂度是O(nlogn)。
通过对数组的前64个元素进行分析(num=63),我们发现数组呈现一定的规律,不断重复,如下图所示:

0 
1 
1 2 
1 2 2 3 
1 2 2 3 2 3 3 4 
1 2 2 3 2 3 3 4 2 3 3 4 3 4 4 5 
1 2 2 3 2 3 3 4 2 3 3 4 3 4 4 5 2 3 3 4 3 4 4 5 3 4 4 5 4 5 5 6 

由此我们发现0112是一个基础元素,不断循环反复,可以推论:如果已知第一个元素是result[0],那么第二第三个元素为result[0]+1,第四个元素为result[0]+2,由此获得前4个元素result[0]~result[3];以这4个元素为基础,我们可以得到
result[4]=result[0]+1,result[5]=result[1]+1…,
result[8]=result[0]+1,result[9]=result[1]+1… ,
result[12]=result[0]+2,result[13]=result[1]+2…;
以此类推可以获得全部的数组。

Java版代码1:

public class Solution {
    public int[] countBits(int num) {
        int[] result = new int[num + 1];
        int range = 1;
        result[0] = 0;
        boolean stop = false;
        while (!stop) {
            stop = fillNum(result, range);
            range *= 4;
        }
        return result;
    }

    public boolean fillNum(int[] nums, int range) {
        for (int i = 0; i < range; i++) {
            if (range + i < nums.length) {
                nums[range + i] = nums[i] + 1;
            } else {
                return true;
            }
            if (2 * range + i < nums.length) {
                nums[2 * range + i] = nums[i] + 1;
            }
            if (3 * range + i < nums.length) {
                nums[3 * range + i] = nums[i] + 2;
            }
        }
        return false;
    }
}

Java版代码2:

public class Solution {
    public int[] countBits(int num) {
        int[] result=new int[num+1];
        result[0]=0;
        for(int i=1;i<=num;i++){
            result[i]=getCount(i);
        }
        return result;
    }
    public int getCount(int num){
        int count=0;
        while(num!=0){
            if((num&1)==1){
                count++;
            }
            num/=2;
        }
        return count;
    }
}

Leet Code OJ 338. Counting Bits [Difficulty: Easy]

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原文地址:http://blog.csdn.net/lnho2015/article/details/50924299

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