1356. Sort Integers by The Number of 1 Bits.md

1356. Sort Integers by The Number of 1 Bits

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leetcode-cn Daily Challenge on November 6th, 2020.

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Difficulty : Easy

Related Topics : Sort、Bit Manipulation

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Given an integer array arr. You have to sort the integers in the array in ascending order by the number of 1's in their binary representation and in case of two or more integers have the same number of 1's you have to sort them in ascending order.

Return the sorted array.

Example 1:

Input: arr = [0,1,2,3,4,5,6,7,8]
Output: [0,1,2,4,8,3,5,6,7]
Explantion: [0] is the only integer with 0 bits.
[1,2,4,8] all have 1 bit.
[3,5,6] have 2 bits.
[7] has 3 bits.
The sorted array by bits is [0,1,2,4,8,3,5,6,7]

Example 2:

Input: arr = [1024,512,256,128,64,32,16,8,4,2,1]
Output: [1,2,4,8,16,32,64,128,256,512,1024]
Explantion: All integers have 1 bit in the binary representation, you should just sort them in ascending order.

Example 3:

Input: arr = [10000,10000]
Output: [10000,10000]

Example 4:

Input: arr = [2,3,5,7,11,13,17,19]
Output: [2,3,5,17,7,11,13,19]

Example 5:

Input: arr = [10,100,1000,10000]
Output: [10,100,10000,1000]

Constraints:

• 1 <= arr.length <= 500
• 0 <= arr[i] <= 10^4

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Solution

• mine
  • Java
    • Runtime: 7 ms, faster than 84.93%, Memory Usage: 39.6 MB, less than 5.73% of Java online submissions

      // O(N*logN)time
      // O(N)space
      public int[] sortByBits(int[] arr) {
          Integer[] res = new Integer[arr.length];
          for (int i = 0; i < arr.length; i++) {
              res[i] = arr[i];
          }
          Arrays.sort(res, (o1, o2) -> {
              int c1 = getCount(o1);
              int c2 = getCount(o2);
              if (c1 == c2) {
                  return o1 - o2;
              }
              return c1 - c2;
          });
          for (int i = 0; i < arr.length; i++) {
              arr[i] = res[i];
          }
          return arr;
      }
      
      int getCount(int num) {
          int i = 0;
          while (num > 0) {
              i++;
              num -= num & (-num);
          }
          return i;
      }
      

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• the most votes

• Runtime: 2 ms, faster than 99.86%, Memory Usage: 39.5 MB, less than 5.73% of Java online submissions

  // O(N*logN)time
  // O(N)space
  public int[] sortByBits(int[] arr) {
      int[] ans = new int[arr.length];
      for (int i = 0; i < arr.length; i++) {
          //Integer.bitCount(arr[i])转化为二进制一的数量
          ans[i] = Integer.bitCount(arr[i]) * 10000000 + arr[i];
      }
      Arrays.sort(ans);
      for (int i = 0; i < ans.length; i++) {
          ans[i] = ans[i] % 10000000;
      }
      return ans;
  }
  

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