我目前正在查看两个非常大的Peak
对象列表,通过重写equals
方法并循环遍历这两个列表,比较每个峰值与其他峰值。有没有更有效的方法来完成这个任务?我的列表可以包含约10,000个元素,这意味着最多10000 * 10000次比较。
我的峰值对象的代码:
public class Peak extends Object{
private final SimpleIntegerProperty peakStart;
private final SimpleIntegerProperty peakEnd;
private final SimpleIntegerProperty peakMaxima;
private final SimpleIntegerProperty peakHeight;
private final SimpleIntegerProperty peakWidth;
private final SimpleStringProperty rname;
public Peak(int peakStart, int peakEnd, int peakMaxima, int peakHeight, String rname) {
this.peakStart = new SimpleIntegerProperty(peakStart);
this.peakEnd = new SimpleIntegerProperty(peakEnd);
this.peakMaxima = new SimpleIntegerProperty(peakMaxima);
this.peakHeight = new SimpleIntegerProperty(peakHeight);
this.peakWidth = new SimpleIntegerProperty(peakEnd - peakStart);
this.rname = new SimpleStringProperty(rname);
}
public String getRname() {
return rname.get();
}
public SimpleStringProperty rnameProperty() {
return rname;
}
public int getPeakWidth() {
return peakWidth.get();
}
public int getPeakHeight() {
return peakHeight.get();
}
public int getPeakStart() {
return peakStart.get();
}
public int getPeakEnd() {
return peakEnd.get();
}
public int getPeakMaxima() {
return peakMaxima.get();
}
@Override
public String toString() {
return "Peak{" +
"peakStart= " + peakStart.get() +
", peakEnd= " + peakEnd.get() +
", peakHeight= " + peakHeight.get() +
", rname= " + rname.get() +
'}';
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Peak peak = (Peak) o;
if (!peakMaxima.equals(peak.peakMaxima)) return false;
return rname.equals(peak.rname);
}
@Override
public int hashCode() {
int result = peakMaxima.hashCode();
result = 31 * result + rname.hashCode();
return result;
}
}
以下是比较对象的循环代码。
List<Peak> interestingPeaks = new ArrayList<>();
if(peakListOne != null && peakListTwo != null){
for(Peak peak : peakListOne){
for(Peak peak2 : peakListTwo){
if(peak.equals(peak2)){ //number one, check the rnames match
if((peak2.getPeakHeight() / peak.getPeakHeight() >= 9) || (peak.getPeakHeight() / peak2.getPeakHeight() >= 9)){
interestingPeaks.add(peak);
}
}
}
}
}
return interestingPeaks;
这段代码主要是匹配最大值的位置和字符串 rname
,然后将峰值添加到 interestingPeaks
列表中,如果其中一个峰值的高度是另一个峰值的9倍以上。