leetcode-面试题40-最小的k个数

题目描述：

 

 方法一：快排 O(N)   java

class Solution {
    public int[] getLeastNumbers(int[] arr, int k) {
        if(k == 0|| arr.length == 0){
            return new int[0];
        }
        return quickSerch(arr,0,arr.length -1,k-1);

    }
    private int[] quickSerch(int[] nums,int lo,int hi,int k){
        int r = partition(nums,lo,hi);
        if(r == k){
            return Arrays.copyOf(nums,r+1);
        }
        return r > k? quickSerch(nums,lo,r - 1,k): quickSerch(nums,r+1,hi,k);
    }

    private int partition(int[] nums,int lo,int hi){
        int pivot = nums[lo];
        int l = lo+1,r = hi;
        while(l <= r){
            if (nums[l] > pivot && nums[r] < pivot){
                int temp = nums[l];
                nums[l] = nums[r];
                nums[r] = temp;
            }
            if (nums[l] <= pivot) l++;
            if (nums[r] >= pivot) r--;
        }
        nums[lo] = nums[r];
        nums[r] = pivot;
        return r;
    }
}

方法二：大根堆：O(nlogk)  小根堆：O(nlogn)

class Solution {
    public int[] getLeastNumbers(int[] arr, int k) {
        if (k == 0 || arr.length == 0){
            return new int[0];
        }
        Queue<Integer> pq = new PriorityQueue<>((v1,v2) -> v2 - v1);
        for (int num:arr){
            if (pq.size() < k){
                pq.offer(num);
            }else if(num < pq.peek()){
                pq.poll();
                pq.offer(num);
            }
        }
        int[] res = new int[pq.size()];
        int idx = 0;
        for(int num:pq){
            res[idx++] = num;
        }
        return res;
    }
}

方法三：二叉搜索树O(nlogk)

class Solution {
    public int[] getLeastNumbers(int[] arr, int k) {
        if (k == 0|| arr.length == 0){
            return new int[0];
        }

        TreeMap <Integer,Integer> map = new TreeMap<>();
        int cnt = 0;
        for (int num: arr) {
            if (cnt < k) {
                map.put(num, map.getOrDefault(num, 0) + 1);
                cnt++;
                continue;
            } 

            Map.Entry<Integer,Integer> entry = map.lastEntry();
            if (entry.getKey() > num){
                map.put(num,map.getOrDefault(num,0) + 1);
                if (entry.getValue() == 1){
                    map.pollLastEntry();
                }else{
                    map.put(entry.getKey(),entry.getValue() - 1);
                }
            }
        }
        int [] res = new int[k];
        int idx = 0;
        for (Map.Entry<Integer,Integer> entry:map.entrySet()){
            int freq = entry.getValue();
            while(freq-- > 0){
                res[idx++] = entry.getKey();
            }
        }
        return res;
    }
}

方法四：计数排序 O(N)

class Solution {
    public int[] getLeastNumbers(int[] arr, int k) {
        if (k == 0|| arr.length == 0){
            return new int[0];
        }
        int[] counter = new int[10001];
        for(int num: arr){
            counter[num]++;
        }
        int[] res = new int[k];
        int idx = 0;
        for (int num = 0; num < counter.length;num++){
            while(counter[num]-- > 0 && idx < k){
                res[idx++] = num;
            }
            if(idx == k){
                break;
            }
        }
        return res;
    }
}