未排序数组中第 K 个最小/最大元素| 系列 1 · GeeksForGeeks 中文系列教程

# 未排序数组中第 K 个最小/最大元素| 系列 1 > 原文： [https://www.geeksforgeeks.org/kth-smallestlargest-element-unsorted-array/](https://www.geeksforgeeks.org/kth-smallestlargest-element-unsorted-array/) 给定一个数组和一个数`k`（其中`k`小于数组的大小），我们需要找到给定数组中第`k`个最小的元素。假定有 11 个数组元素是不同的。 **示例**： ``` 输入: arr [] = {7, 10, 4, 4, 3, 20, 15} k = 3 输出: 7 输入: arr [] = {7, 10, 4, 3, 20, 15} k = 4 输出: 10 ``` 我们已经讨论了类似的[问题来打印 k 个最大元素](https://www.geeksforgeeks.org/k-largestor-smallest-elements-in-an-array/)。 **方法 1（简单解决方案）** 一个简单的解决方案是使用`O(N log N)`排序算法对给定数组进行排序，例如[合并排序](http://geeksquiz.com/merge-sort/)， [堆排序](http://geeksquiz.com/heap-sort/)等，并返回已排序数组中索引`k-1`处的元素。该解决方案的时间复杂度为`O(N log N)` ## C++ ```cpp // Simple C++ program to find k'th smallest element #include <algorithm> #include <iostream> using namespace std; // Function to return k'th smallest element in a given array int kthSmallest(int arr[], int n, int k) { // Sort the given array sort(arr, arr + n); // Return k'th element in the sorted array return arr[k - 1]; } // Driver program to test above methods int main() { int arr[] = { 12, 3, 5, 7, 19 }; int n = sizeof(arr) / sizeof(arr[0]), k = 2; cout << "K'th smallest element is " << kthSmallest(arr, n, k); return 0; } ``` ## Java ```java // Java code for kth smallest element // in an array import java.util.Arrays; import java.util.Collections; class GFG { // Function to return k'th smallest // element in a given array public static int kthSmallest(Integer[] arr, int k) { // Sort the given array Arrays.sort(arr); // Return k'th element in // the sorted array return arr[k - 1]; } // driver program public static void main(String[] args) { Integer arr[] = new Integer[] { 12, 3, 5, 7, 19 }; int k = 2; System.out.print("K'th smallest element is " + kthSmallest(arr, k)); } } // This code is contributed by Chhavi ``` ## Python3 ```py # Python3 program to find k'th smallest # element # Function to return k'th smallest # element in a given array def kthSmallest(arr, n, k): # Sort the given array arr.sort() # Return k'th element in the # sorted array return arr[k-1] # Driver code if __name__=='__main__': arr = [12, 3, 5, 7, 19] n = len(arr) k = 2 print("K'th smallest element is", kthSmallest(arr, n, k)) # This code is contributed by # Shrikant13 ``` ## C# ```cs // C# code for kth smallest element // in an array using System; class GFG { // Function to return k'th smallest // element in a given array public static int kthSmallest(int[] arr, int k) { // Sort the given array Array.Sort(arr); // Return k'th element in // the sorted array return arr[k - 1]; } // driver program public static void Main() { int[] arr = new int[] { 12, 3, 5, 7, 19 }; int k = 2; Console.Write("K'th smallest element" + " is " + kthSmallest(arr, k)); } } // This code is contributed by nitin mittal. ``` ## PHP ```php <?php // Simple PHP program to find // k'th smallest element // Function to return k'th smallest // element in a given array function kthSmallest($arr, $n, $k) { // Sort the given array sort($arr); // Return k'th element // in the sorted array return $arr[$k - 1]; } // Driver Code $arr = array(12, 3, 5, 7, 19); $n =count($arr); $k = 2; echo "K'th smallest element is ", kthSmallest($arr, $n, $k); // This code is contributed by anuj_67\. ?> ``` **输出**： ``` K'th smallest element is 5 ``` **方法 2（使用最小堆 – HeapSelect）** 我们发现时间复杂度中的第`k`个最小元素要好于`O(N log N)`。一个简单的优化方法是创建给定`n`个元素的[最小堆](http://geeksquiz.com/binary-heap/)，并调用`k`次`extractMin()`。以下是上述方法的 C++ 实现。 ``` // A C++ program to find k'th smallest element using min heap #include <climits> #include <iostream> using namespace std; // Prototype of a utility function to swap two integers void swap(int* x, int* y); // A class for Min Heap class MinHeap { int* harr; // pointer to array of elements in heap int capacity; // maximum possible size of min heap int heap_size; // Current number of elements in min heap public: MinHeap(int a[], int size); // Constructor void MinHeapify(int i); // To minheapify subtree rooted with index i int parent(int i) { return (i - 1) / 2; } int left(int i) { return (2 * i + 1); } int right(int i) { return (2 * i + 2); } int extractMin(); // extracts root (minimum) element int getMin() { return harr[0]; } // Returns minimum }; MinHeap::MinHeap(int a[], int size) { heap_size = size; harr = a; // store address of array int i = (heap_size - 1) / 2; while (i >= 0) { MinHeapify(i); i--; } } // Method to remove minimum element (or root) from min heap int MinHeap::extractMin() { if (heap_size == 0) return INT_MAX; // Store the minimum vakue. int root = harr[0]; // If there are more than 1 items, move the last item to root // and call heapify. if (heap_size > 1) { harr[0] = harr[heap_size - 1]; MinHeapify(0); } heap_size--; return root; } // A recursive method to heapify a subtree with root at given index // This method assumes that the subtrees are already heapified void MinHeap::MinHeapify(int i) { int l = left(i); int r = right(i); int smallest = i; if (l < heap_size && harr[l] < harr[i]) smallest = l; if (r < heap_size && harr[r] < harr[smallest]) smallest = r; if (smallest != i) { swap(&harr[i], &harr[smallest]); MinHeapify(smallest); } } // A utility function to swap two elements void swap(int* x, int* y) { int temp = *x; *x = *y; *y = temp; } // Function to return k'th smallest element in a given array int kthSmallest(int arr[], int n, int k) { // Build a heap of n elements: O(n) time MinHeap mh(arr, n); // Do extract min (k-1) times for (int i = 0; i < k - 1; i++) mh.extractMin(); // Return root return mh.getMin(); } // Driver program to test above methods int main() { int arr[] = { 12, 3, 5, 7, 19 }; int n = sizeof(arr) / sizeof(arr[0]), k = 2; cout << "K'th smallest element is " << kthSmallest(arr, n, k); return 0; } ``` **输出**： ``` K'th smallest element is 5 ``` 该解决方案的时间复杂度为`O(n + kLogn)`。 **方法 3（使用最大堆）** 我们也可以使用最大堆来找到第`k`个最小元素。以下是算法。 1）建立给定数组的前`k`个元素（`arr[0]`至`arr[k-1]`）的最大堆 MH 2）对于每个元素，在第`k`个元素之后（`arr[k]`至`arr[n-1]`），将其与 MH 的根进行比较。 ……a）如果元素小于根，则将其设为根并为 MH 调用建堆 ……b）否则将其忽略。 //步骤 2 为`O((n-k) * logk)` 3）最后，MH 的根是第`k`个最小元素。该解决方案的时间复杂度为`O(k + (n-k) * Logk)` 以下是上述算法的 C++ 实现 ``` // A C++ program to find k'th smallest element using max heap #include <climits> #include <iostream> using namespace std; // Prototype of a utility function to swap two integers void swap(int* x, int* y); // A class for Max Heap class MaxHeap { int* harr; // pointer to array of elements in heap int capacity; // maximum possible size of max heap int heap_size; // Current number of elements in max heap public: MaxHeap(int a[], int size); // Constructor void maxHeapify(int i); // To maxHeapify subtree rooted with index i int parent(int i) { return (i - 1) / 2; } int left(int i) { return (2 * i + 1); } int right(int i) { return (2 * i + 2); } int extractMax(); // extracts root (maximum) element int getMax() { return harr[0]; } // Returns maximum // to replace root with new node x and heapify() new root void replaceMax(int x) { harr[0] = x; maxHeapify(0); } }; MaxHeap::MaxHeap(int a[], int size) { heap_size = size; harr = a; // store address of array int i = (heap_size - 1) / 2; while (i >= 0) { maxHeapify(i); i--; } } // Method to remove maximum element (or root) from max heap int MaxHeap::extractMax() { if (heap_size == 0) return INT_MAX; // Store the maximum vakue. int root = harr[0]; // If there are more than 1 items, move the last item to root // and call heapify. if (heap_size > 1) { harr[0] = harr[heap_size - 1]; maxHeapify(0); } heap_size--; return root; } // A recursive method to heapify a subtree with root at given index // This method assumes that the subtrees are already heapified void MaxHeap::maxHeapify(int i) { int l = left(i); int r = right(i); int largest = i; if (l < heap_size && harr[l] > harr[i]) largest = l; if (r < heap_size && harr[r] > harr[largest]) largest = r; if (largest != i) { swap(&harr[i], &harr[largest]); maxHeapify(largest); } } // A utility function to swap two elements void swap(int* x, int* y) { int temp = *x; *x = *y; *y = temp; } // Function to return k'th largest element in a given array int kthSmallest(int arr[], int n, int k) { // Build a heap of first k elements: O(k) time MaxHeap mh(arr, k); // Process remaining n-k elements. If current element is // smaller than root, replace root with current element for (int i = k; i < n; i++) if (arr[i] < mh.getMax()) mh.replaceMax(arr[i]); // Return root return mh.getMax(); } // Driver program to test above methods int main() { int arr[] = { 12, 3, 5, 7, 19 }; int n = sizeof(arr) / sizeof(arr[0]), k = 4; cout << "K'th smallest element is " << kthSmallest(arr, n, k); return 0; } ``` **输出**： ``` K'th smallest element is 12 ``` **方法 4（快速选择）** 如果在第一步中将[快速排序](http://geeksquiz.com/quick-sort/)用作排序算法，则这是对方法 1 的优化。在快速排序中，我们选择一个枢轴元素，然后将枢轴元素移动到正确的位置并围绕该数组对其进行分区。这个想法不是要进行完整的快速排序，而是停在枢轴本身是第`k`个最小元素的位置。同样，不要针对枢轴的左侧和右侧重复出现，而是根据枢轴的位置针对其中之一进行重复播放。该方法最坏的情况是时间复杂度为 `O(n^2)`，但平均而言，它的工作时间为`O(n)`。 ## C++ ``` #include <climits> #include <iostream> using namespace std; int partition(int arr[], int l, int r); // This function returns k'th smallest element in arr[l..r] using // QuickSort based method. ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT int kthSmallest(int arr[], int l, int r, int k) { // If k is smaller than number of elements in array if (k > 0 && k <= r - l + 1) { // Partition the array around last element and get // position of pivot element in sorted array int pos = partition(arr, l, r); // If position is same as k if (pos - l == k - 1) return arr[pos]; if (pos - l > k - 1) // If position is more, recur for left subarray return kthSmallest(arr, l, pos - 1, k); // Else recur for right subarray return kthSmallest(arr, pos + 1, r, k - pos + l - 1); } // If k is more than number of elements in array return INT_MAX; } void swap(int* a, int* b) { int temp = *a; *a = *b; *b = temp; } // Standard partition process of QuickSort(). It considers the last // element as pivot and moves all smaller element to left of it // and greater elements to right int partition(int arr[], int l, int r) { int x = arr[r], i = l; for (int j = l; j <= r - 1; j++) { if (arr[j] <= x) { swap(&arr[i], &arr[j]); i++; } } swap(&arr[i], &arr[r]); return i; } // Driver program to test above methods int main() { int arr[] = { 12, 3, 5, 7, 4, 19, 26 }; int n = sizeof(arr) / sizeof(arr[0]), k = 3; cout << "K'th smallest element is " << kthSmallest(arr, 0, n - 1, k); return 0; } ```