leetcode [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.

Example 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [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.

题目大意：

计算[0,n]这个区间内所有整数中1的个数。

解法：

最简单的做法可以对这个区间的每个数字求解1的个数，但是这样做的话，时间复杂度比较大。

java:

class Solution {
    public int[] countBits(int num) {
        int[] res=new int[num+1];
        for (int i=0;i<=num;i++){
            int n=i;
            while(n!=0){
                n&=(n-1);
                res[i]++;
            }
        }
        return res;
    }
}

还可以使用动态规划的方法，这种做法真机智，直接res[i]=res[i>>1]+(1&i），这样比起上面时间复杂度小了很多，只有O（n）。

class Solution {
    public int[] countBits(int num) {
        int[] res=new int[num+1];
        for (int i=1;i<=num;i++){
            res[i]=res[i>>1]+(1&i);
        }
        return res;
    }
}