堆排及应用例子

最小化堆堆排
堆的难点在于：

1. 取出最小值时，需要向下过滤，找出次小值顶替最小值位置

2. 插入时，需要向上回溯找到插入值的位置

    public class Heap {
    
        private int[] nums;
        private int base = 0;//堆容量基数，扩容时使用
    
        //capacity 初始化堆大小
        public Heap (int capacity){//创建一个空堆
            nums = new int[capacity+1];//使用nums[0]来存放当前堆中元素个数
            base = capacity;
        }
    
        //堆扩容
        private void dilatation () {
            int[] newHeap = new int[2*base + 1];
    
            for (int i=0; i < nums.length; i++){
                newHeap[i] = nums[i];
            }
            nums = newHeap;
        }
    
        //向上回溯（核心）
        private void adjustUp (int[] nums, int poi){
            int lf = nums[2*poi];//左子
            if (lf > nums[poi]) {
                nums[2*poi] = nums[poi];
                nums[poi] = lf;
            }
    
            if (2*poi+1 <= nums[0]) {//存在右子（右子可能不存在）
                int rt = nums[2 * poi + 1];//右子
                if (rt > nums[poi]) {
                    nums[2*poi+1] = nums[poi];
                    nums[poi] = rt;
                }
    
            }
        }
    
        //一轮向上过滤
        private void up (int[] nums){
            if (nums[0] <= 1) return;
            int poi = nums[0]/2;
    
            while (poi > 0) {
                adjustUp(nums, poi);
                poi--;
            }
    
        }
    
        //堆排
        private void sortHeap (int[] arr) {
    
            int len = arr.length;
            if (len == 0) return;
    
            int[] nums = new int[len+1];
            for (int i=1; i < len+1; i++)
                nums[i] = arr[i-1];
            nums[0] = len;
    
            int count = nums[0];
            while (nums[0]-1 > 0){
                up(nums);
    
                //每轮排序完都应将最堆顶（最大值）与堆尾交换
                int t = nums[0];
                int tmp = nums[t];
                nums[t] = nums[1];
                nums[1] = tmp;
    
                nums[0]--;
    
            }
            nums[0] = count;
    
            for (int i=0; i < len; i++)
                arr[i] = nums[i+1];
    
        }
    
        //堆排
        private void sortHeap () {
    
            int count = nums[0];
            while (nums[0]-1 > 0){
                up(nums);
    
                //每轮排序完都应将最堆顶（最大值）与堆尾交换
                int t = nums[0];
                int tmp = nums[t];
                nums[t] = nums[1];
                nums[1] = tmp;
    
                nums[0]--;
    
            }
            nums[0] = count;
    
    
        }
    
    
        //入堆（可能需要扩容）: 堆的构建是通过入堆时产生的，所以没有构造函数
        public void put (int num) {
            int poi = nums[0] + 1;
            if (poi == nums.length)//数组已满
                dilatation();//扩容
    
            nums[poi] = num;
            nums[0]++;
            sortHeap();
    
        }
    
        //遍历
        public void travelHeap () {
            int len = nums[0];
            if (len == 0) return;
    
            for (int i=1; i <= len; i++){
                System.out.print(nums[i] + " ");
            }
            System.out.println();
        }
    
    
        //最小元素出堆
        public int pop (){
            if (nums[0] == 0) return -1;//堆空
            int target = nums[1];
    
            //重新调整 (此处如果用指针可以将时间复杂度变为o(1))
            if (nums[0] > 1) {
                //整体前移一位即可
                for (int i=2; i <= nums[0]; i++)
                    nums[i-1] = nums[i];
            }
    
            nums[0]--;
            return target;
        }
    
        public static void main (String[] args) {
            int[] arr = {1, 8, 1, 3, 8, 3, 8, 4, 5, 1};
            Heap heap1 = new Heap(10);
            heap1.sortHeap(arr);
    
            for (int i : arr)
                System.out.print(i + " ");
    
            System.out.println();
    
            Heap heap2 = new Heap(10);
            heap2.put(10);
            heap2.put(1);
            heap2.put(8);
            heap2.put(12);
            heap2.travelHeap();
    
            while (heap2.pop() != -1)
             heap2.travelHeap();
    
        }
    
    }