# 1356. Sort Integers by The Number of 1 Bits
###### tags: `Leetcode` `Easy` `Bit Manipulation`
Link: https://leetcode.com/problems/sort-integers-by-the-number-of-1-bits/
## 思路
先用```Integer.bitCount()```算出每个数字的the number of 1 bits
然后排序
## Code

class Solution {
    public int[] sortByBits(int[] arr) {
        int n = arr.length;
        int[][] cnt = new int[n][2];
        for(int i=0; i<n; i++){
            int num = arr[i];
            cnt[i][0] = num;
            cnt[i][1] = Integer.bitCount(num);
        }
        Arrays.sort(cnt, new Comparator<int[]>(){
           public int compare(int[] a, int[] b){
               if(a[1] != b[1]) {
                    return a[1]-b[1];
                } else {
                    return a[0]-b[0];
                }
           } 
        });
        
        for(int i=0; i<n; i++) arr[i] = cnt[i][0];
        return arr;
    }
}

c++

class Solution {
public:
    static int count1s(int num){
        int cnt = 0;
        while(num){
            cnt += num&1;
            num = num >> 1;
        }
        return cnt;
    }
    static bool compare(int a, int b){
        int n1 = count1s(a);
        int n2 = count1s(b);
        if(n1==n2) return a<b;
        else return n1<n2;
    }
    vector<int> sortByBits(vector<int>& arr) {
        sort(arr.begin(), arr.end(), compare);
        return arr;
    }
};