好吧,我承认我闲的蛋疼
问题:3000万条的记录取最大的前50条数据
这题的时间复杂度应该是3000W*lg50
需要取前50个数构造heap恒长为50的最小堆
第51个数开始,与根节点比较,如果大于根节点,与根节点交换,并进行一次最小堆的minHeapify过程
内存2G很宽裕
代码如下(PS:我是没时间做3000W的数据,除非我真的蛋疼)
package org.leon.testcase;
public class Main {
public static void main(String[] args) {
int[] target = {12,4,3,56,11,122,131,312,2,3321,3,4,13,1,231,3,4,5,6,5,34,3};
int[] ary = new int[5];
int j=0;
MinHeap heap = new MinHeap();
for (int i = 0; i < target.length; i++) {
if(j<5){
ary[j]=target[i];
j++;
continue;
}
if(j==5){
heap.builtMinHeap(ary);
}
if(ary[0]<target[i]){
ary[0]=target[i];
heap.minHeapify(ary,0);
}
}
for (int i = 0; i < ary.length; i++) {
System.out.println(ary[i]);
}
}
}
class MinHeap {
@SuppressWarnings("unused")
private int parent(int i) {
if (i == 0)
return -1;
return i / 2;
}
private int left(int i) {
if (i == 0)
return 1;
return 2 * i;
}
private int right(int i) {
if (i == 0)
return 2;
return 2 * i + 1;
}
public void minHeapify(int[] ary, int i) {
int l = left(i);
int r = right(i);
int min = i;
if (l < ary.length) {
if (ary[l] < ary[i]) {
min = l;
}
}
if (r < ary.length) {
if (ary[r] < ary[min]) {
min = r;
}
}
if (min != i) {
int temp = ary[i];
ary[i] = ary[min];
ary[min] = temp;
minHeapify(ary, min);
}
}
public void builtMinHeap(int[] a) {
for (int i = (a.length - 1) / 2; i >= 0; i--) {
minHeapify(a, i);
}
}
}