java - 如何使用 Java 查找 ArrayList<Entry> 中的最小值和最大值

Question

我正在尝试找到最小值和最大值

ArrayList<Entry>

例如我的 ArrayList 看起来像这样:

ArrayList<Entry> test = new ArrayList<Entry>();
test.add(new Entry(20, 0));
test.add(new Entry(5, 0));
test.add(new Entry(15, 0));

现在我想要这个列表的最小值(5)和最大值(20)。

我尝试过:

Collections.min(test);

但它说:

Bound mismatch: The generic method min(Collection<? extends T>) of type Collections is not applicable for the arguments (ArrayList<Entry>). The inferred type Entry is not a valid substitute for the bounded parameter <T extends Object & Comparable<? super T>>

我也尝试过:

test.length()

所以我可以做一个for循环。但对于这种 ArrayList 也失败了。

Answer 1

首先，定义一个Comparator<Entry>它定义了 Entry 的排序:

class EntryComparator implements Comparator<Entry> {
  @Override public int compare(Entry a, Entry b) {
    // ... whatever logic to compare entries.
    // Must return a negative number if a is "less than" b
    // Must return zero if a is "equal to" b
    // Must return a positive number if a is "greater than" b
  }
}

然后遍历列表，将每个元素与当前的最小和最大元素进行比较:

Comparator<Entry> comparator = new EntryComparator();
Iterator<Entry> it = list.iterator();
Entry min, max;
// Assumes that the list is not empty
// (in which case min and max aren't defined anyway).

// Any element in the list is an upper bound on the min
// and a lower bound on the max.
min = max = it.next();

// Go through all of the other elements...
while (it.hasNext()) {
  Entry next = it.next();
  if (comparator.compare(next, min) < 0) {
    // Next is "less than" the current min, so take it as the new min.
    min = next;
  }
  if (comparator.compare(next, max) > 0) {
    // Next is "greater than" the current max, so take it as the new max.
    max = next;
  }
}