问题陈述：找到给定字符串中字母字符的重复（仅来自az的所有小写）并从最低到最高排序。 如果两个字符具有相同的重复次数，则具有较大ASCII值的字符被认为较小。
虽然问题很简单，但我试图使用Comparator对最终答案进行排序，而不是使用我自己的排序。 这是我做的：
private static void importantString(String context) { HashMap<Character, Integer> importance = new HashMap<Character, Integer>(); String alpha="abcdefghijklmnopqrstuvwxyz"; for(int i=0; i<26; i++){ importance.put(alpha.charAt(i),0); } for(int i=0; i<context.length(); i++){ char temp = context.charAt(i); Integer val = importance.get(temp); importance.put(temp,++val); } //To sort ArrayList<Map.Entry<Character, Integer>> l = new ArrayList(importance.entrySet()); Collections.sort(l, new Comparator<Map.Entry<Character, Integer>>(){ public int compare(Map.Entry<Character, Integer> o1, Map.Entry<Character, Integer> o2) { if (o1.getValue()==o2.getValue()){ if (o1.getKey() > o2.getKey()){ return -1; } else{ return 1; } } else { return o1.getValue().compareTo(o2.getValue()); } }}); System.out.println(l); for (Entry<Character, Integer> m: l){ System.out.print(m.getKey()+" "); } System.out.println(); }现在，这个definitley适用于较小的测试用例。 例如，
但是，我有一个非常大的测试用例，这是我得到的（我的排序数组是这个，因为测试用例字符串非常大）：
[r=38083, p=38223, v=38223, f=38268, e=38269, u=38306, z=38320, k=38341, g=38342, c=38396, o=38418, q=38418, b=38422, n=38467, x=38476, y=38477, l=38525, m=38534, w=38575, d=38580, a=38619, s=38648, t=38653, h=38787, j=38791, i=38839]注意，o = 38418，q = 38418。 在这种情况下，'q'的优先级应低于o，因为它具有更高的ASCII值。 但它没有反映出来。
对于像oooqqq这样的小型测试用例，我确实得到了正确的结果。 任何解释为什么？
Problem statement: Find the repetition of the alphabet characters in a given string (all lower case from a-z only) and sort them from lowest to highest. If two character have the same number of repetitions, the character with the greater ASCII value is considered to be smaller.
Although the problem is easy, I was trying to use Comparator to sort the final answer instead of using my own sorting. Here is what I did:
private static void importantString(String context) { HashMap<Character, Integer> importance = new HashMap<Character, Integer>(); String alpha="abcdefghijklmnopqrstuvwxyz"; for(int i=0; i<26; i++){ importance.put(alpha.charAt(i),0); } for(int i=0; i<context.length(); i++){ char temp = context.charAt(i); Integer val = importance.get(temp); importance.put(temp,++val); } //To sort ArrayList<Map.Entry<Character, Integer>> l = new ArrayList(importance.entrySet()); Collections.sort(l, new Comparator<Map.Entry<Character, Integer>>(){ public int compare(Map.Entry<Character, Integer> o1, Map.Entry<Character, Integer> o2) { if (o1.getValue()==o2.getValue()){ if (o1.getKey() > o2.getKey()){ return -1; } else{ return 1; } } else { return o1.getValue().compareTo(o2.getValue()); } }}); System.out.println(l); for (Entry<Character, Integer> m: l){ System.out.print(m.getKey()+" "); } System.out.println(); }Now, this definitley works for smaller test cases. For example,
However, I had a very large test case this is what I get (my sorted array is this since the test case string is very large):
[r=38083, p=38223, v=38223, f=38268, e=38269, u=38306, z=38320, k=38341, g=38342, c=38396, o=38418, q=38418, b=38422, n=38467, x=38476, y=38477, l=38525, m=38534, w=38575, d=38580, a=38619, s=38648, t=38653, h=38787, j=38791, i=38839]Notice, o=38418 and q=38418. In this case 'q' should have a lower precedence than o since it has a higher ASCII value. Yet it does not reflect.
For smaller test cases like oooqqq, I do get the correct results. Any explainations why?
最满意答案
你的问题是这一行：
if (o1.getValue()==o2.getValue()){你在这里使用引用相等。 由于==两边都是Integer类型，因此它们会进行参考相等性测试，但仅对于值-128 <= value <= 127那些Integer被保证为相同的对象（请参阅用于自动装箱的Integer.valueOf这里： importance.put(temp,++val); ）
你可以简单地用int值的比较来替换它：
if (o1.getValue().intValue()==o2.getValue().intValue()){此外，您还可以使用Integer.compareTo和Character.compareTo重写该方法：
public int compare(Map.Entry<Character, Integer> o1, Map.Entry<Character, Integer> o2) { int res = o1.getValue().compareTo(o2.getValue()); return res == 0 ? o2.getKey().compareTo(o1.getKey()) : res; }Your issue is this line:
if (o1.getValue()==o2.getValue()){You're using reference equality here. Since both sides of == are of type Integers they are tested for reference equality, but only for values -128 <= value <= 127 those Integers are guarantied to be the same objects (see Integer.valueOf which is used for autoboxing here: importance.put(temp,++val);)
You could simply replace this with a comparison of the int values:
if (o1.getValue().intValue()==o2.getValue().intValue()){Furthermore you could also rewrite the method by using Integer.compareTo and Character.compareTo:
public int compare(Map.Entry<Character, Integer> o1, Map.Entry<Character, Integer> o2) { int res = o1.getValue().compareTo(o2.getValue()); return res == 0 ? o2.getKey().compareTo(o1.getKey()) : res; }更多推荐
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