需要帮助按第二个元素然后按第一个元素（Java）对二维数组进行排序

Question

I have a two dimensional array that is holding the value of 5 cards. The first element of each of the 5 arrays represents the suit of the card, and the second element represents the card value.

I want to sort the 2d array by the second element, and then by the first element while maintaining the sorted order of the second element (if that makes sense). For example all suits of ones will be lower on the sorted list than all suits of two. So for example, {{0,1},{2,1},{0,2}} should become {{0,1},{2,1},{0,2}}.

Here is what I have:

 // {{3,2}, {2,2}, {0,1}, {1,0}, {2,3}} should become 
 // {{1,0}, {0,1}, {2,2}, {3,2}, {2,3}}

 int[][] hand = {{3,2},{2,2},{0,1},{1,0},{2,3}};
 sort(hand);

 public static void sort(int[][] hand){
    Arrays.sort(hand, new Comparator<int[]>(){
        public int compare(int[] o1, int[] o2){
            return Integer.valueOf(o1[1]).compareTo(Integer.valueOf(o2[1]));
        }
    });
 }

This is outputting {{1,0},{0,1},{3,2},{2,2},{2,3}}. Does anyone have any suggestions?

Answer 1

Solution 1: sort the arrays by the second element, and then sort the arrays by the first element. Since Arrays.sort is stable, that's equivalent to first comparing by the first element, then the second.

Solution 2: modify your comparator as follows:

Arrays.sort(hand, new Comparator<int[]>() {
  public int compare(int[] o1, int[] o2) {
    if (o1[0] == o2[0]) {
      return Integer.compare(o1[1], o2[1]);
    } else {
      return Integer.compare(o1[0], o2[0]);
    }
  }
});

or, with Guava (disclosure: I contribute to Guava), you can just write the comparator as

  public int compare(int[] o1, int[] o2) {
    return ComparisonChain.start()
      .compare(o1[0], o2[0])
      .compare(o1[1], o2[1])
      .result();
  }

Answer 2

Would this work for you:

int compare1 = Integer.valueOf(o1[1]).compareTo(Integer.valueOf(o2[1]);
if(compare1 != 0)
    return compare1;
else
    return Integer.valueOf(o1[0]).compareTo(Integer.valueOf(o2[0]));

Answer 3

I will add a Java 8 solution, just in case someone wants to know.

Arrays.sort(hand, (o1, o2) -> o1[1] == o2[1] ? Integer.compare(o1[0], o2[0]) 
                                             : Integer.compare(o1[1], o2[1]));

Basically, it compares the first element of each array when second element is equal, otherwise just compare the second elements directly.

Answer 4

您可以像这样以更紧凑的方式使用 Java 8 Comparator

Arrays.sort(hand, Comparator.comparing(x -> ((int[])x)[1]).thenComparing(x -> ((int[])x)[0]));