Sort Integers by The Number Of 1 Bits
You are given an integer array arr. 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 array after sorting it.
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.
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 104
思路一：先统计数字的 bit 位数，然后用排序
public int[] sortByBits(int[] arr) {
Map<Integer, Integer> map = new HashMap<>();
for (int i : arr) {
map.put(i, Integer.bitCount(i));
}
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr.length - 1; j++) {
int c1 = map.get(arr[j]);
int c2 = map.get(arr[j + 1]);
if (c1 > c2 || (c1 == c2 && arr[j] > arr[j + 1])) {
int t = arr[j];
arr[j] = arr[j + 1];
arr[j + 1] = t;
}
}
}
return arr;
}