哈希集包含的字符位置准确，但字符位置不正确

Question

我有一个类，它确定2个字符数组中有多少匹配字符。使用HashSetcontains方法，如果一个字符数组包含第二个数组中的Character，则显示该字符
问题是，如果两个匹配的字符出现在多个位置。

例如，如果array 1 = adcd和array 2 = a05ddd，则d出现3次，而不是2次。
如何修改此项以获得正确的字符数？代码在应该生成"add"时生成"addd"对于不正确的字符，结果将是"a--dd"

HashSet<Character> hash = new HashSet<Character>();        
String word1 = "adcd";
String word2 = "a05ddd";
char[] ch1 = word1.toCharArray();
char[] ch2 = word2.toCharArray();
Character character = null;
String charLocation = "";
int count = 0;
for (int i = 0; i < ch1.length; i++) 
{
hash.add(ch1[i]); 

}
for (int i = 0; i < ch2.length; i++) 
{
character = ch2[i]; 
if (hash.contains(character)) 
{
charLocation = charLocation + character;
count++;
} 
if (!hashSet.contains(character)) 
correctCharPlacements = correctCharPlacements + "-";
}

Answer 1

很可能需要收集有关字符频率的数据，因此应该用映射替换set。然后，可以根据常见字符和这两个单词中字符的最小频率来构建结果字符串：

// helper method to build frequency map for a word/string
private static Map<Character, Integer> freqMap(String word) {
return word.chars()
.mapToObj(c -> (char)c)
.collect(Collectors.toMap(x -> x, x -> 1, Integer::sum, LinkedHashMap::new));
}
static String common(String w1, String w2) {
if (null == w1 || null == w2) {
return null;
}
Map<Character, Integer> map1 = freqMap(w1);
Map<Character, Integer> map2 = freqMap(w2);
Set<Character> commonChars = new LinkedHashSet<>(map1.keySet());
commonChars.retainAll(map2.keySet());
return commonChars.stream()
.map(x -> String.valueOf(x).repeat(Math.min(map1.get(x), map2.get(x))))
.collect(Collectors.joining(""));
}

测试：

System.out.println("common = " + common("adcd", "a05ddd"));

输出：

common = add

---

更新

由于String::repeat从Java 11开始就可用，因此可以使用String::join+Collections.nCopies:将其替换为Java 8兼容代码

static String common(String w1, String w2) {
if (null == w1 || null == w2) {
return null;
}
Map<Character, Integer> map1 = freqMap(w1);
Map<Character, Integer> map2 = freqMap(w2);
Set<Character> commonChars = new LinkedHashSet<>(map1.keySet());
commonChars.retainAll(map2.keySet());

return commonChars.stream()
.map(x -> String.join("", Collections.nCopies(Math.min(map1.get(x), map2.get(x)), String.valueOf(x))))
.collect(Collectors.joining(""));
}

在线演示：

System.out.println("Java version: " + System.getProperty("java.version"));
System.out.println("common = " + common("adcd", "a05ddd"));

输出：

Java version: 1.8.0_201
common = add

---

更新2

为了保持结果中第二个字符串中字符的顺序，并用'-'替换缺失的字符，可以实现以下方法：

static String common2(String w1, String w2) {
if (null == w1 || null == w2) {
return null;
}
Map<Character, Integer> map1 = freqMap(w1);
StringBuilder sb = new StringBuilder();
for (char c : w2.toCharArray()) {
if (map1.containsKey(c) && map1.get(c) > 0) {
sb.append(c);
map1.merge(c, -1, Integer::sum); // decrement the character counter in the first map
} else if (!map1.containsKey(c)) { // character missing in the first word
sb.append('-');
}
}
return sb.toString();
}

测试

String[][] tests = {
{"adcd", "ad05dd"},
{"adcd", "d05add"},
{"adcd", "d05dadd"},
};
for (String[] test : tests) {
System.out.printf("common(%s, %s) = %s%n", test[0], test[1], common(test[0], test[1]));
}

输出：

common(adcd, ad05dd) = ad--d
common(adcd, d05add) = d--ad
common(adcd, d05dadd) = d--da