面试题 40. 最小的 k 个数

输入：arr = [3,2,1], k = 2
输出：[1,2] 或者 [2,1]

输入：arr = [0,1,2,1], k = 1
输出：[0]

import heapq

class Solution:
    def getLeastNumbers(self, arr: List[int], k: int) -> List[int]:
        if k == 0:
            return []
        heap = []
        for e in arr:
            heapq.heappush(heap, e)
        return heapq.nsmallest(k, heap)

class Solution {
    public int[] getLeastNumbers(int[] arr, int k) {
        if (k == 0) {
            return new int[]{};
        }
        PriorityQueue<Integer> bigRoot = new PriorityQueue<>(k, Collections.reverseOrder());
        for (int e : arr) {
            if (bigRoot.size() < k) {
                bigRoot.offer(e);
            } else if (e < bigRoot.peek()) {
                bigRoot.poll();
                bigRoot.offer(e);
            }
        }
        int[] res = new int[k];
        for (int i = 0; i < k; ++i) {
            res[i] = bigRoot.poll();
        }
        return res;
    }
}

/**
 * @param {number[]} arr
 * @param {number} k
 * @return {number[]}
 */
var getLeastNumbers = function (arr, k) {
  // 排序
  // return arr.sort((a,b)=>a-b).slice(0,k)
  // ==========================================
  // 快排思想
  let left = 0;
  let right = arr.length - 1;
  while (left < right) {
    let i = partition(left, right);
    if (i <= k) {
      left = i + 1;
    }
    if (i >= k) {
      right = i - 1;
    }
  }
  function partition(left, right) {
    let pivot = arr[left];
    while (left < right) {
      while (left < right && arr[right] >= pivot) {
        right--;
      }
      arr[left] = arr[right];
      while (left < right && arr[left] <= pivot) {
        left++;
      }
      arr[right] = arr[left];
    }
    arr[left] = pivot;
    return left;
  }
  return arr.slice(0, k);
};

class Solution {
public:
    int partition(vector<int>& arr, int begin, int end) {
        int l = begin;
        int r = end;
        int povit = arr[begin];

        while (l < r) {
            // 从右边开始，找到第一个小于povit的数字（用于交换）
            while (l < r && arr[r] >= povit) { r--; }
            while (l < r && arr[l] <= povit) { l++; }
            if (l < r) { swap(arr[l], arr[r]); }
        }

        swap(arr[begin], arr[l]);
        return l;
    }

    void partSort(vector<int>& arr, int begin, int end, int target) {
        if (begin >= end) {
            return;
        }

        // 思路类似快排，这样做比堆排序时间复杂度低
        // C++中，stl提供partial_sort()方法，就是这种实现方式
        int mid = partition(arr, begin, end);
        if (mid == target) {
            return;
        } else if (target < mid) {
            partSort(arr, begin, mid - 1, target);
        } else {
            partSort(arr, mid + 1, end, target);
        }

        return;
    }

    vector<int> getLeastNumbers(vector<int>& arr, int k) {
        partSort(arr, 0, arr.size() - 1, k - 1);
        vector<int> ret(arr.begin(), arr.begin() + k);
        return ret;
    }
};