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Counting Bits-LeetCode#338

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.

思路:题目意为给定一个非负整数,计算每个从0至num的数字二进制中1的个数,返回数组。用动态规划的思想可以举例

dp[0] = 0;

dp[1] = dp[0] + 1;

dp[2] = dp[0] + 1;

dp[3] = dp[1] +1;

dp[4] = dp[0] + 1;

dp[5] = dp[1] + 1;

dp[6] = dp[2] + 1;

dp[7] = dp[3] + 1;

dp[8] = dp[0] + 1;

等于

dp[0] = 0;

dp[1] = dp[1-1] + 1;

dp[2] = dp[2-2] + 1;

dp[3] = dp[3-2] +1;

dp[4] = dp[4-4] + 1;

dp[5] = dp[5-4] + 1;

dp[6] = dp[6-4] + 1;

dp[7] = dp[7-4] + 1;

dp[8] = dp[8-8] + 1;

能得出dp[index] = dp[index – offset] + 1;

代码如下:

public class CountingBits {
    public static int[] countBits(int num) {
        int[] res = new int[num + 1];
        int i = 1;
        int offset = 1;
        while (i < num + 1) {
            if (offset * 2 == i) {
                offset = offset * 2;
            }
            res[i] = res[i - offset] + 1;
            i++;
        }
        return res;
    }
    
}

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