Codewars题记 ：Count the number of Duplicates

1、题目：

Count the number of Duplicates
Write a function that will return the count of distinct case-insensitive alphabetic characters and numeric digits that occur more than once in the input string. The input string can be assumed to contain only alphabets (both uppercase and lowercase) and numeric digits.

Example
"abcde" -> 0 # no characters repeats more than once
"aabbcde" -> 2 # 'a' and 'b'
"aabBcde" -> 2 # 'a' occurs twice and 'b' twice (`b` and `B`)
"indivisibility" -> 1 # 'i' occurs six times
"Indivisibilities" -> 2 # 'i' occurs seven times and 's' occurs twice
"aA11" -> 2 # 'a' and '1'
"ABBA" -> 2 # 'A' and 'B' each occur twice

2、我的解决方案

 1 class CountingDuplicates {
 2     public static int duplicateCount(String text) {
 3         
 4         Map<Character, Integer > map = new HashMap<Character, Integer>();
 5         
 6         for (Character one : text.toLowerCase().toCharArray()) {
 7             if (map.containsKey(one)) {
 8                 Integer sum = map.get(one);
 9                 sum++;
10                 map.put(one, sum);
11             } else {
12                 map.put(one, 1);
13             }
14         }
15         
16         int result = 0;
17         
18         for (Entry<Character, Integer> entry : map.entrySet()) {
19             if (entry.getValue()>1) {
20                 result++;
21             }
22         }
23         
24         return result;
25     }
26 }

3、最佳解决方案

public class CountingDuplicates {
  public static int duplicateCount(String text) {
    int ans = 0;
    text = text.toLowerCase();
    while (text.length() > 0) {
      String firstLetter = text.substring(0,1);
      text = text.substring(1);
      if (text.contains(firstLetter)) ans ++;
      text = text.replace(firstLetter, "");
    }
    return ans;
  }
}

4、总结
　　最佳解决方案每次将一个字符截取出来然后查找，查找完后就会把字符串中这个字符删除掉，再进行下一轮查找。