count sort

principle

Determine the position of the element by counting the number of occurrences of each element.

Application scenarios

It is suitable for the case where the element value is small and there are many repetitions. The essence of counting sorting is sorting by counting the number of times, which has better performance than the comparison sorting algorithm. When the applicable scenarios are limited, such as sorting company employees by age, etc.

performance

The time complexity is only O(n), and the space complexity is O(n).

code

    /* 非常巧妙的实现 */
    public static void countSort(int a[]){
        if(a == null)
            return;
        int max = a[0];
        for(int i = 1; i < a.length; i++){ //求当前数组最大值
            if(max < a[i])
                max = a[i];
        }

        int[] count = new int[max+1];
        for(int i = 0; i < count.length; i++){ //初始化
            count[i] = 0;
        }
        for(int i = 0; i < a.length; i++){ //统计a中各个元素出现的次数
            count[a[i]]++;
        }
        int k = 0;
        for(int i = 0; i <= max; i++){
            for(int j = 0; j < count[i]; j++){
                //值i在数组a中出现了count[i]次，就将a[]设置count[i]次，相当于给数组a[]排序
                a[k++] = i;
            }
        }
    }

    /* 按算法导论第三版实现 */
    public static int[] countSort1(int a[]){
        if(a == null)
            return null;
        int max = a[0];
        for(int i = 1; i < a.length; i++){ //求当前数组最大值
            if(max < a[i])
                max = a[i];
        }

        int[] count = new int[max+1];
        for(int i = 0; i < count.length; i++){ //初始化
            count[i] = 0;
        }
        for(int i = 0; i < a.length; i++){ //统计a中各个元素出现的次数
            count[a[i]]++;
        }
        for(int i = 1; i < count.length; i++){ //累计次数
            count[i] += count[i-1];
        }

        int[] b = new int[a.length];
        for(int i = a.length-1; i >= 0; i--){ //通过统计的次数决定元素位置 
            b[count[a[i]] - 1] = a[i]; //注意索引从0开始
            count[a[i]]--; //减一是为了防止相同元素插入同一位置
        }
        return b;
    }

    public static void main(String[] args){
        int[] a = new int[]{6, 7, 8, 10, 6, 7, 6, 6, 16, 10};
        int[] b = countSort1(a);
        for(int v: b)
            System.out.println(v);
    }

references

[1] The sword refers to the offer
[2] The third edition of Introduction to Algorithms