[算法]打印N个数组的整体最大Top K
2016-02-14 17:24
302 查看
题目:
有N个长度不一的数组,所有的数组都是有序的,请从大到小打印这N个数组整体最大的前K个数。例如:
输入含有N行元素的二维数组代表N个一维数组。
219,405,538,845,971
148,558
52,99,348,691
再输入整数K=5,则打印:
Top 5:971,845,961,558,538。
要求:
1.如果所有数组的元素个数小于K,则从小到大打印所有的数。2.时间复杂度为O(KlogN)。
解答:
利用堆结构和堆排序完成。import java.util.Arrays;
public class Problem_20_PrintMaxTopK {
public static class HeapNode {
public int value;
public int arrNum;
public int index;
public HeapNode(int value, int arrNum, int index) {
this.value = value;
this.arrNum = arrNum;
this.index = index;
}
}
public static void printTopK(int[][] matrix, int topK) {
int heapSize = matrix.length;
HeapNode[] heap = new HeapNode[heapSize];
for (int i = 0; i != heapSize; i++) {
int index = matrix[i].length - 1;
heap[i] = new HeapNode(matrix[i][index], i, index);
heapInsert(heap, i);
}
System.out.println("TOP " + topK + " : ");
for (int i = 0; i != topK; i++) {
if (heapSize == 0) {
break;
}
System.out.print(heap[0].value + " ");
if (heap[0].index != 0) {
heap[0].value = matrix[heap[0].arrNum][--heap[0].index];
} else {
swap(heap, 0, --heapSize);
}
heapify(heap, 0, heapSize);
}
}
public static void heapInsert(HeapNode[] heap, int index) {
while (index != 0) {
int parent = (index - 1) / 2;
if (heap[parent].value < heap[index].value) {
swap(heap, parent, index);
index = parent;
} else {
break;
}
}
}
public static void heapify(HeapNode[] heap, int index, int heapSize) {
int left = index * 2 + 1;
int right = index * 2 + 2;
int largest = index;
while (left < heapSize) {
if (heap[left].value > heap[index].value) {
largest = left;
}
if (right < heapSize && heap[right].value > heap[largest].value) {
largest = right;
}
if (largest != index) {
swap(heap, largest, index);
} else {
break;
}
index = largest;
left = index * 2 + 1;
right = index * 2 + 2;
}
}
public static void swap(HeapNode[] heap, int index1, int index2) {
HeapNode tmp = heap[index1];
heap[index1] = heap[index2];
heap[index2] = tmp;
}
public static int[][] generateRandomMatrix(int maxRow, int maxCol,
int maxValue) {
if (maxRow < 0 || maxCol < 0) {
return null;
}
int[][] matrix = new int[(int) (Math.random() * maxRow) + 1][];
for (int i = 0; i != matrix.length; i++) {
matrix[i] = new int[(int) (Math.random() * maxCol) + 1];
for (int j = 0; j != matrix[i].length; j++) {
matrix[i][j] = (int) (Math.random() * maxValue);
}
Arrays.sort(matrix[i]);
}
return matrix;
}
public static void printMatrix(int[][] matrix) {
for (int i = 0; i != matrix.length; i++) {
for (int j = 0; j != matrix[i].length; j++) {
System.out.print(matrix[i][j] + " ");
}
System.out.println();
}
}
public static void main(String[] args) {
int[][] matrix = generateRandomMatrix(5, 10, 1000);
printMatrix(matrix);
System.out.println("===========================");
printTopK(matrix, 100);
}
}
相关文章推荐
- UVa--679 Dropping Balls(模拟)
- 大规模网站架构技术原理透析(1)
- Linux GCC常用命令
- 老杨聊架构:每个架构师都应该研究下康威定律
- nginx referer限制
- Zabbix高可用,实现zabbix的无缝切换,无故障时间
- JavaWeb学习总结(二)——Tomcat服务器学习和使用(一)
- centos6中安装svn1.8版本
- Kafka Zero-Copy 使用分析
- shell升级完整记录
- 那都是别人的架构
- Transformer架构解析
- Hadoop 源码目录树
- zabbix Server端在CentOS6.5上的安装过程
- centos设置ssh命令无密码登录
- 查找linux系统下的端口被占用进程的两种方法
- OpenGL坐标系
- Hadoop之MapReduce自定义二次排序流程实例详解
- CentOS 6.5 64位下安装MySQL 5.7.11
- shell升级