java - 字串中的字数

Question

我有一个特殊情况(希望保持较低的时间复杂度)我需要计算可以从给定字符串创建单词的次数。我下面的代码只设法得到它一次，如果可以多次创建单词失败。有什么想法吗？

//String that needs to be searched
String s= "ccoptarra";

//Words that need to be found
String[] words = { "car", "pot", "bag" };

ArrayList count = new ArrayList();
HashMap map = new HashMap();
for (int i = 0; i < words.length; i++) {
    count.add(0);
    String w = words[i];
    map.put(w, "");
    for (int j = 0; j < w.length(); j++) {
        if (s.contains(String.valueOf(w.charAt(j)))) {
            map.put(w, map.get(w).toString() + w.charAt(j));
            if (map.get(w).equals(w))
                count.add(i, ((int)count.get(i)) + 1);
        }
    }
}
for (int i = 0; i < count.size(); i++)
    System.out.println("Word: " + words[i] + ", count = " + count.get(i));

输出:

Word: car, count = 1
Word: pot, count = 1
Word: bag, count = 0

Answer 1

您可以使用Map来保存每个字符和字符数。之后，只需循环测试单词并从 map 中为该单词选择最少的字符数。这是代码

        //String that needs to be searched
    String s= "ccoptarra";

    //Words that need to be found
    String[] words = {"car","pot","bag"};

    List<Integer> count = new ArrayList<Integer>();
    Map<Character, Integer> map = new HashMap<Character, Integer>();
    for (int i=0; i<s.length();i++) {
        char c = s.charAt(i);
        if (map.get(c) != null) {
            map.put(c, map.get(c) + 1);
        } else {
            map.put(c, 1);
        }
    }

    for (int i=0; i<words.length; i++) {
        int min = Integer.MAX_VALUE;
        for (int j=0; j<words[i].length(); j++) {
            Integer value = map.get(words[i].charAt(j));
            if (value == null) {
                min = 0;
            } else if (value < min) {
                min = value;
            }
        }
        count.add(min == Integer.MAX_VALUE ? 0 : min);
    }

    for(int i=0; i<count.size(); i++)
        System.out.println("Word: "+words[i]+", count = "+count.get(i));