有效地比较Java中的两个对象列表

Question

I'm currently looking through two very large lists of Peak Objects, by overriding the equals method and looping through the two lists, comparing every peak to every other peak. Is there a more efficient way of doing this? My lists can be ~10,000 elements, which means up to 10000 * 10000 comparisons.

The code for my peak object:

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;
}
}

And my loop for comparing the objects is here.

 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;

The code is basically matching the position of the maxima, and the rname , which is just a String. Then appending the peak to the interestingPeaks list if the height of one is a factor of 9x larger than the other.

Answer 1

Appreciate that if the two lists were sorted by maxima and name, you could simply make a single linear pass down both lists, and compare items side by side. If the two lists were in fact completely equal, then you would never find a pair from the two lists which were not equal.

List<Peak> p1;
List<Peak> p2;

p1.sort((p1, p2) -> {
    int comp = Integer.compare(p1.getPeakMaxima(), p2.getPeakMaxima());
    return comp != 0 ? comp : p1.getRname().compareTo(p2.getRname());
});

// and also sort the second list

Now we can just walk down both lists and check for a comparison failure:

for (int i=0; i < p1.size(); ++i) {
    if (!p1.get(i).equals(p2.get(i))) {
        System.out.println("peaks are not equal");
        break;
    }
}

This reduces an O(N^2) operation to one which is O(N*lgN) , which is the penalty for doing both sorts (the final walk down the list is O(N) , and would be negligible with either approach).