题目：

有很多随机大小的明信片，也有很多随机大小的信封。希望把一个明信片装到多个信封里。明信片只能装入比自己大的信封，信封也只能装入更大的信封。相同的大小无法装入。

为了保证最大数量的信封被使用和装入最多数量的卡片，请设计算法。

解决方法

1.将明信片和信封各自排序。

2.找到第一个明信片。

3.找到没使用的第一个信封。

4. 是否可装入，是则装入。否按顺序向前从大到小找卡片，能装入则装入。不能装入则继续从大到小找卡片。直到没有卡片可找。

代码

package sort;import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;public class NestedEnvelope {public static void main(String[] args) {List<Card> cards = new ArrayList<>();for (int i = 0; i < 5; ++i) {Card card = new Card();card.size = i * 2;card.id = i;cards.add(card);}List<Envelop> evelops = new ArrayList<>();for (int i = 0; i < 10; ++i) {Envelop env = new Envelop();env.size = i;env.id = i;evelops.add(env);}HashMap<Card, List<Envelop>> results = putInCard(cards, evelops);for (Map.Entry<Card, List<Envelop>> entry : results.entrySet()) {System.out.println("[");System.out.println("card:" + entry.getKey().size);for (Envelop env : entry.getValue()) {System.out.println("envelop:" + env.size);}System.out.println("]");}}static class Card implements Comparable<Card> {int id;int size;@Overridepublic int compareTo(Card o) {// TODO Auto-generated method stubreturn size - o.size;}}static class Envelop implements Comparable<Envelop> {int id;int size;@Overridepublic int compareTo(Envelop o) {// TODO Auto-generated method stubreturn size - o.size;}}public static HashMap<Card, List<Envelop>> putInCard(List<Card> cards, List<Envelop> envelops) {Collections.sort(cards);Collections.sort(envelops);HashMap<Card, List<Envelop>> cardWithEnvelops = new HashMap<Card, List<Envelop>>();for (int i = 0; i < cards.size(); ++i) {Card card = cards.get(i);Card nextCard = null;if (i + 1 < cards.size()) {nextCard = cards.get(i + 1);}List<Envelop> cardEnvelops = new ArrayList<Envelop>();cardWithEnvelops.put(card, cardEnvelops);Envelop lastEnvelop = null;while (envelops.size() > 0) {Envelop envelop = envelops.get(0);if ((lastEnvelop != null && envelop.size == lastEnvelop.size) || envelop.size == card.size) {for (int indexCard = i - 1; indexCard >= 0; --indexCard) {List<Envelop> preEnvelops = cardWithEnvelops.get(cards.get(indexCard));if (preEnvelops.size() == 0) {preEnvelops.add(envelop);} else {Envelop maxEnv = preEnvelops.get(preEnvelops.size() - 1);if (maxEnv.size < envelop.size) {preEnvelops.add(envelop);}}}envelops.remove(envelop);continue;}if (envelop.size > card.size && (nextCard == null || envelop.size <= nextCard.size)) {cardEnvelops.add(envelop);lastEnvelop = envelop;envelops.remove(envelop);} else {break;}}}return cardWithEnvelops;}
}

打印：

[
card:4
envelop:5
envelop:6
]
[
card:8
envelop:9
]
[
card:0
envelop:1
envelop:2
]
[
card:2
envelop:3
envelop:4
]
[
card:6
envelop:7
envelop:8
]

总结：

结果并不是只有一种装法最多，但是设计的这个算法能保证装的最多卡片，和最多信封。