【Leetcode】Bulls and Cows

题目链接：https://leetcode.com/problems/bulls-and-cows/

题目：

You are playing the following Bulls and Cows game with your friend: You write down a number and
ask your friend to guess what the number is. Each time your friend makes a guess, you provide a hint that indicates how many digits in said guess match your secret number exactly in both digit and position (called "bulls") and how many digits match the secret
number but locate in the wrong position (called "cows"). Your friend will use successive guesses and hints to eventually derive the secret number.

For example:

Secret number:  "1807"
Friend‘s guess: "7810"

Hint: 1 bull
and 3 cows.
(The bull is 8,
the cows are 0, 1 and 7.)

Write a function to return a hint according to the secret number and friend‘s guess, use A to
indicate the bulls and B to indicate the cows. In the above example, your function should
return "1A3B".

Please note that both secret number and friend‘s guess may contain duplicate digits, for example:

Secret number:  "1123"
Friend‘s guess: "0111"

In this case, the 1st 1 in
friend‘s guess is a bull, the 2nd or 3rd 1 is
a cow, and your function should return "1A1B".

You may assume that the secret number and your friend‘s guess only contain digits, and their lengths are always equal.

思路：

本题属于hashmap tag，我考虑问题的时候无脑强行用了hashmap，结果代码写的很长，不过思路很清晰也简单，网上有更简洁的解法。

用两个hashmap保存guess和secret字符情况，先求bull，后对剩下的字符比较求cow。

算法：

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1. public String getHint(String secret, String guess) {
2. String hint = "";
3. int bull = 0, cow = 0;
4. HashMap<Character, Integer> map = new HashMap<Character, Integer>();
5. HashMap<Character, Integer> gmap = new HashMap<Character, Integer>();
6. for (int i = 0; i < secret.length(); i++) {
7. char ssecret = secret.charAt(i);
8. char cguess = guess.charAt(i);
9. if (map.containsKey(ssecret)) {
10. map.put(ssecret, map.get(ssecret) + 1);
11. } else {
12. map.put(ssecret, 1);
13. }
14. if (gmap.containsKey(cguess)) {
15. gmap.put(cguess, gmap.get(cguess) + 1);
16. } else {
17. gmap.put(cguess, 1);
18. }
19. }
20. for (int i = 0; i < secret.length(); i++) {// 求bull
21. char ssecret = secret.charAt(i);
22. char cguess = guess.charAt(i);
23. if (ssecret == cguess) {
24. bull++;
25. map.put(ssecret, map.get(ssecret) - 1);
26. gmap.put(ssecret, gmap.get(ssecret) - 1);
27. }
28. }
29. Iterator<Character> it = map.keySet().iterator();
30. while (it.hasNext()) { //求cow，对secret中非bull字符判断是否存在于guess剩下字符中
31. char c = it.next();
32. while (map.get(c) > 0) { // 如果c是非bull字符
33. if (gmap.containsKey(c) && gmap.get(c) > 0) {//如果guess剩下还有c字符
34. cow++;
35. map.put(c, map.get(c) - 1);
36. gmap.put(c, gmap.get(c) - 1);
37. } else {
38. break;
39. }
40. }
41. }
42. return hint + bull + "A" + cow + "B";
43. }