[8,4,5] (one smaller on first dimension)
[9,2,5] (one smaller on second dimension)
[9,4,1] (one smaller on third dimension)
[0,4,5], [8,2,5], [8,4,1]
[9,2,5], [9,4,1], [0,4,5], [8,2,5], [8,4,1]
[8,2,5], [9,2,1].
[9,4,1], [0,4,5], [8,2,5], [8,4,1], [9,2,1]
注意 [8,2,5] 再次出现。请勿重复。
这次的最大和是 [8,2,5]。相邻元素有
[0,2,5], [8,2,1]
此时您应该已经有了想法。
使用最大堆来处理候选集。然后查找具有最大总和的元组需要 O(log |C|) 的时间，其中 C 是候选集。
这个集合可以有多大？这是一个有趣的问题。我会让你自己思考一下。对于您示例中的三个输入集，它是
|C| = O(|A1||A2| + |A2||A3| + |A1||A3|)
O(log(|A1||A2| + |A2||A3| + |A1||A3|))
import java.util.Arrays;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.Set;
public class SortedProduct {
final SortedTuple [] tuples;
final NoDupHeap candidates = new NoDupHeap();
SortedProduct(SortedTuple [] tuple) {
this.tuples = Arrays.copyOf(tuple, tuple.length);
reset();
}
static class SortedTuple {
final int [] elts;
SortedTuple(int... elts) {
this.elts = Arrays.copyOf(elts, elts.length);
Arrays.sort(this.elts);
}
@Override
public String toString() {
return Arrays.toString(elts);
}
}
class RefTuple {
final int [] refs;
final int sum;
RefTuple(int [] index, int sum) {
this.refs = index;
this.sum = sum;
}
RefTuple getSuccessor(int i) {
if (refs[i] == 0) return null;
int [] newRefs = Arrays.copyOf(this.refs, this.refs.length);
int j = newRefs[i]--;
return new RefTuple(newRefs, sum - tuples[i].elts[j] + tuples[i].elts[j - 1]);
}
int [] getTuple() {
int [] val = new int[refs.length];
for (int i = 0; i < refs.length; ++i)
val[i] = tuples[i].elts[refs[i]];
return val;
}
@Override
public int hashCode() {
return Arrays.hashCode(refs);
}
@Override
public boolean equals(Object o) {
if (o instanceof RefTuple) {
RefTuple t = (RefTuple) o;
return Arrays.equals(refs, t.refs);
}
return false;
}
}
RefTuple getInitialCandidate() {
int [] index = new int[tuples.length];
int sum = 0;
for (int j = 0; j < index.length; ++j)
sum += tuples[j].elts[index[j] = tuples[j].elts.length - 1];
return new RefTuple(index, sum);
}
final void reset() {
candidates.clear();
candidates.add(getInitialCandidate());
}
int [] getNext() {
if (candidates.isEmpty()) return null;
RefTuple next = candidates.poll();
for (int i = 0; i < tuples.length; ++i) {
RefTuple successor = next.getSuccessor(i);
if (successor != null) candidates.add(successor);
}
return next.getTuple();
}
/** A max heap of indirect ref tuples that ignores addition of duplicates. */
static class NoDupHeap {
final PriorityQueue<RefTuple> heap =
new PriorityQueue<>((a, b) -> Integer.compare(b.sum, a.sum));
final Set<RefTuple> set = new HashSet<>();
void add(RefTuple t) {
if (set.contains(t)) return;
heap.add(t);
set.add(t);
}
RefTuple poll() {
RefTuple t = heap.poll();
set.remove(t);
return t;
}
boolean isEmpty() {
return heap.isEmpty();
}
void clear() {
heap.clear();
set.clear();
}
}
public static void main(String [] args) {
SortedTuple [] tuples = {
new SortedTuple(9, 8, 0),
new SortedTuple(4, 2),
new SortedTuple(5, 1),
};
SortedProduct product = new SortedProduct(tuples);
for (;;) {
int[] next = product.getNext();
if (next == null) break;
System.out.println(Arrays.toString(next));
}
}
}
这里是一些Python代码。（效率不高 - 直接生成整个列表然后排序可能更好。）
#! /usr/bin/env python
import heapq
def decreasing_tuple_order(*lists):
# Each priority queue element will be:
# (-sum, indices, incrementing_index, sliced)
# The top element will have the largest sum.
if 0 < min((len(l) for l in lists)):
indices = [0 for l in lists]
sliced = [lists[i][indices[i]] for i in range(len(indices))]
queue = [(-sum(sliced), indices, 0, sliced)]
while 0 < len(queue):
#print(queue)
(_, indices, indexable, sliced) = heapq.heappop(queue)
yield sliced
# Can we increment this index?
if indices[indexable] + 1 < len(lists[indexable]):
new_indices = indices[:]
new_indices[indexable] = indices[indexable] + 1
sliced = [lists[i][new_indices[i]] for i in range(len(indices))]
heapq.heappush(queue, (-sum(sliced), new_indices, indexable, sliced))
# Start indexing the next index?
while indexable + 1 < len(lists):
indexable = indexable + 1
if 1 < len(lists[indexable]):
# Start incrementing here.
indices[indexable] = 1
sliced = [lists[i][indices[i]] for i in range(len(indices))]
heapq.heappush(queue, (-sum(sliced), indices, indexable, sliced))
a1 = [9, 8, 0]
a2 = [4, 2]
a3 = [5, 1]
for x in decreasing_tuple_order(a1, a2, a3):
print((x,sum(x)))