topic description
Given a set of playing cards, there are 5 cards in total. Each card has four suits, which are hearts, spades, red squares, and black twists. The value of each card is in the range of [A, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K]. Now given a set of judgment logic, find out its maximum card type.
- Card Type 1: If five cards appear consecutively and have the same suit, it is called a straight flush . Note that when A is included, only two flush combinations of A, 1, 2, 3, 4 and 10, J, Q, K, A are allowed, and similar combinations of Q, K, A, 1, 2 are not allowed.
- Card Type 2: If four cards of the same size appear, it is called a quadruple .
- Card Type Three: If the five cards are of the same suit, it is called a flush .
- Card type 4: If there are five consecutive cards, it is called a straight [tip: It is required that the suits of the five cards cannot be the same, otherwise it is a straight flush ].
- Card Type Five: If there are three cards of the same size and two other cards of the same size, it is called a triple .
- Card Type Six: If three cards of the same size appear and two other cards of a different value, it is called a triple .
- Card type seven: other combinations.
Now use V to represent hearts, B to represent spades, N to represent diamonds, and M to represent twists. Now enter 5 lines of strings, each line of strings contains two character strings separated by spaces. For example:
Input:
1 N
2 N
3 N
4 N
5 N
Output:
1
It is required to analyze according to the input, and output the maximum card face of the combination [the smaller the card type, the larger the card face], the card face is represented by 1...7 numbers. For example, card type one is represented by 1.
topic analysis
public class Main {
public static void main(String[] args) {
p3()
}
static void resolve() {
Map<String, Integer> map = new HashMap<>();
map.put("J", 11);
map.put("Q", 12);
map.put("K", 13);
Map<Integer, List<String>> cardMap = new HashMap<>();
Scanner in = new Scanner(System.in);
boolean hasA = Boolean.FALSE;
for (int i = 0; i < 5; ++i) {
String[] split = in.nextLine().split(" ");
String s1 = split[0];
String s2 = split[1];
if (s1.equals("A")) {
hasA = Boolean.TRUE;
if (cardMap.containsKey(1) && cardMap.containsKey(14)) {
cardMap.get(1).add(s2);
cardMap.get(14).add(s2);
continue;
}
if (cardMap.containsKey(1) && !cardMap.containsKey(14)) {
cardMap.get(1).add(s2);
List<String> list = new ArrayList<>();
list.add(s2);
cardMap.put(14, list);
continue;
}
if (!cardMap.containsKey(1) && cardMap.containsKey(14)) {
cardMap.get(14).add(s2);
List<String> list = new ArrayList<>();
list.add(s2);
cardMap.put(1, list);
continue;
}
List<String> list1 = new ArrayList<>();
list1.add(s2);
List<String> list2 = new ArrayList<>();
list2.add(s2);
cardMap.put(1, list1);
cardMap.put(14, list2);
continue;
}
if (map.containsKey(s1)) {
Integer key = map.get(s1);
if (cardMap.containsKey(key)) {
cardMap.get(key).add(s2);
} else {
List<String> list = new ArrayList<>();
list.add(s2);
cardMap.put(key, list);
}
continue;
}
int key = Integer.parseInt(s1);
if (cardMap.containsKey(key)) {
cardMap.get(key).add(s2);
} else {
List<String> list = new ArrayList<>();
list.add(s2);
cardMap.put(key, list);
}
}
in.close();
// 判断牌型
// 同花顺 5 1 同花 5 4 顺子 5 5 其它 5 7
// 四条 2 2 葫芦 2 3
// 三条 3 6
int res = 7;
int len = cardMap.size();
if (hasA) len -= 1;
if (len == 3) {
res = Math.min(res, 6);
}
if (len == 2) {
for (Map.Entry<Integer, List<String>> item : cardMap.entrySet()) {
int size = item.getValue().size();
if (size == 1 || size == 4) {
res = Math.min(res, 2);
} else {
res = Math.min(res, 3);
}
}
}
if (len == 5) {
Boolean shunzi;
// 判断是不是顺子
List<Integer> collect = cardMap.keySet().stream().sorted().collect(Collectors.toList());
if (hasA) {
List<Integer> integers1 = collect.subList(0, 5);
List<Integer> integers2 = collect.subList(1, 6);
int i1 = integers1.get(4) - integers1.get(0);
int i2 = integers2.get(4) - integers2.get(0);
shunzi = i1 == 4 || i2 == 4;
} else {
int i = collect.get(4) - collect.get(0);
shunzi = i == 4;
}
if (shunzi) res = Math.min(res, 5);
// 判断是不是同花
boolean tonghua = Boolean.FALSE;
Set<String> set = new HashSet<>();
cardMap.values().forEach(set::addAll);
if (set.size() == 1){
tonghua = true;
}
if (tonghua) {
res = Math.min(res, 4);
}
if (tonghua && shunzi) {
res = Math.min(res, 1);
}
}
System.out.println(res);
}
}
It's not too difficult a question, in fact, the analysis is in the code, just take a look and you'll understand.