为什么HashMap的get()在不应该返回时返回null？

Question

我写了一个方法来检查一个字符串是否只有唯一的字符.我发送明显的非唯一字符字符串"11",它返回true而不是false.它发生的原因是get(c)在if (tab.get(c) == null)返回中,null即使该字符'1'已经在HashMap中.

我该怎么做才能获得预期的行为？

/* Check if a string contains only unique characters */
public static boolean isUniqueChars(String s) {

    HashMap<Boolean, Character> tab = new HashMap<Boolean, Character>();
    Character c;

    for (int i = 0; i < s.length(); ++i) {
        c = new Character(s.charAt(i));
        if (tab.get(c) == null)
            tab.put(Boolean.TRUE, c);
        else
            return false;
    }
    return true;
}

public static void main(String[] args) {

    String s = "11";
    System.out.println(isUniqueChars(s));  /* prints true! why?! */
}

Answer 1

你是按角色取物,但地图的关键是Boolean.您希望密钥Character和值是Boolean:

HashMap<Character, Boolean> tab = new HashMap<Character, Boolean>();
Character c;

for (int i = 0; i < s.length(); ++i) {
    c = new Character(s.charAt(i));
    if (tab.get(c) == null)
        tab.put(c, Boolean.TRUE);
    else
        return false;
}
return true;

话说回来:

• 您不需要Character显式创建新的.拳击会为你做到这一点.
• 使用a HashSet<Character>来跟踪你到目前为止看到的角色会更简单.

例如:

Set<Character> set = new HashSet<Character>();
for (int i = 0; i < s.length(); i++) {
    Character c = s.charAt(i);
    // add returns true if the element was added (i.e. it's new) and false
    // otherwise (we've seen this character before)
    if (!set.add(c)) {
        return false;
    }
}
return true;