java 代码实现 Top N 问题

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思路：用小根堆来实现，时间复杂度是O(nlgK),而且内存消耗是 K。时间复杂度消耗在代码中aaa 处
如果用海量数据排序，内存放不下，得用归并排序，最好最坏平均都是 O(nlgn)
代码如下：

package AlgorithmTopK;public class TopK {public int[] createHeapOther(int a[], int k) {int[] result = new int[k];    for (int i = 0; i < k; i++) {      result[i] = a[i];    }    for (int i = 1; i < k; i++) {      int child = i;      int parent = (i - 1) / 2;      int temp = a[i];      while (parent >= 0 &&child!=0&& result[parent] >temp) {        result[child] = result[parent];        child = parent;        parent = (parent - 1) / 2;      }      result[child] = temp;    }    return result;}public int[] createHeap(int input[], int K) {  //创建小根堆,复杂度最坏是nlgKint heap[] = new int[K];for(int i=0;i<K;i++)heap[i] = input[i];for(int i = 1;i < heap.length;i++) {int child = i;int parent = (child-1) / 2;while(parent >= 0 && child!=0 && heap[parent] > heap[child]) {int temp = heap[child];heap[child] = heap[parent];heap[parent] = temp;child = parent;parent = (parent - 1) / 2;}}return heap;}public void insertHeap(int heap[], int value) {heap[0] = value;int parent = 0;while(parent<heap.length) {   //这个循环复杂度最坏是  logKint lchild = parent*2 + 1;int rchild = parent*2 + 2;int minIndex = parent;  //指向左右儿子中最小的if(lchild < heap.length && heap[lchild] < heap[parent])minIndex = lchild;if(rchild < heap.length && heap[rchild] < heap[minIndex])minIndex = rchild;if(minIndex == parent) {break;}else {int temp = heap[minIndex];heap[minIndex] = heap[parent];heap[parent] = temp;parent = minIndex;}}}public int[] getTopKByHeap(int input[], int K) {int result[] = createHeap(input, K);   //复杂度最坏是 O(nlgK)for(int i=K;i<input.length;i++) {if(input[i] > result[0]) insertHeap(result, input[i]);  //复杂度最坏是O(nlgK)，而且内存消耗就K，不然海量数据排序，内存放不下，得用归并排序，最好最坏平均都是 O(nlgn)}return result;}public static void main(String[] args) {int a[] = {4,3,5,1,2,8,9,10};//待找TOP K 的排海量数据Nint result[] = new TopK().getTopKByHeap(a, 6);System.out.print("TOP K is :");for(int resultItem : result) {System.out.print(resultItem + "   ");}}}

结果：

TOP K is :3 4 5 10 9 8

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