338. Counting Bits

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LeetCode

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Problem

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.

Code

如何去掉末尾的 1

1. i - (i & -i) i & -i可以得到仅保留最后一位 1 的数字，例如110 & (-110)得到10
2. i & (i - 1)

class Solution {
    public int[] countBits(int num) {
        int[] dp = new int[num + 1];

        for(int i = 1; i <= num; i++){
            dp[i] = dp[i - (i & -i)] + 1;
        }

        return dp;
    }
}

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

class Solution {
    public int[] countBits(int num) {
        int[] res = new int[num + 1];

        for(int i = 1; i <= num; i++){
            if(i % 2 == 0){
                res[i] = res[i / 2];
            } else {
                res[i] = res[i - 1] + 1;
            }
        }
        return res;
    }
}