1356. Sort Integers by The Number of 1 Bits

题目描述

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.

Example1

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]

Example2

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.

解决思路

使用系统的排序函数，更改比较器即可

class Solution {
    public int[] sortByBits(int[] arr) {
        Map<Integer, Integer> map = new HashMap<>();
        List<Integer> list = new ArrayList<>();
        for (int x:arr) {
            list.add(x);
            map.put(x, getNums(x));
        }

        Collections.sort(list, new Comparator<Integer>(){
            public int compare(Integer num1, Integer num2) {
                if (map.get(num1) != map.get(num2)) {
                    return map.get(num1)-map.get(num2);
                } else {
                    return num1-num2;
                }
            }
        });

        for(int i=0;i<arr.length;i++) {
            arr[i]=list.get(i);
        }

        return arr;
    }

    public int getNums(int x) {
        int res=0;
        while (x!=0) {
            res+=x%2;
            x/=2;
        }

        return res;
    }
}