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LeetCode 411: Minimum Unique Word Abbreviation

时间:2017-10-13 17:54:11      阅读:144      评论:0      收藏:0      [点我收藏+]

标签:one   return   bsp   rate   span   val   word   class   valueof   

Note: This solution is combined with Trie + Generialized Abbreviation:

1. Since it looks for smallest one, we need persist a string to find minimum length.

2. Lots of mistakes:

                                index not moving (index++)

                                index is messed up. (s.charAt(index) - ‘a‘)

                                data is not intialized.

class Solution {
    class TrieNode {
        TrieNode[] children = new TrieNode[26];
        boolean isWord = false;
    }
    
    private TrieNode root;
    private List<String> result;
    private String minL = "";
    public String minAbbreviation(String target, String[] dictionary) {
        if (dictionary.length == 0) return String.valueOf(target.length());
        root = new TrieNode();
        result = new ArrayList<>();
        for (String s : dictionary) addWord(s);
        generateComb(target, "", 0, 0);
        for (String s : result) {
            if (!search(s, root, 0, 0) && (minL.length() == 0 || s.length() < minL.length())) {
                minL = s;
            }
        }
        return minL;
    }
    
    private void addWord(String s) {
        TrieNode current = root;
        for (int i = 0; i < s.length(); i++) {
            if (current.children[s.charAt(i) - ‘a‘] == null) {
                current.children[s.charAt(i) - ‘a‘] = new TrieNode();
            }
            current = current.children[s.charAt(i) - ‘a‘];
        }
        current.isWord = true;
    }
    
    private boolean search(String s, TrieNode root, int index, int num) {
        if (root == null) return false;
        if (num > 0) {
            for (int i = 0; i < 26; i++) {
                if (search(s, root.children[i], index, num - 1)) return true;
            }
            return false;
        }
        if (index == s.length()) return root.isWord;
        if (Character.isDigit(s.charAt(index))) {
            int current = 0;
            while (index < s.length() && Character.isDigit(s.charAt(index))) {
                current = current * 10 + (s.charAt(index++) - ‘0‘);
            }
            return search(s, root, index, current);
        }
        return search(s, root.children[s.charAt(index) - ‘a‘], index + 1, num);
    }
    
    private void generateComb(String word, String current, int count, int index) {
        if (index == word.length()) {
            if (count > 0) {
                current += String.valueOf(count);
            }
            result.add(current);
            return;
        }
        
        generateComb(word, current, count + 1, index + 1);
        generateComb(word, current + (count > 0 ? count : "") + word.charAt(index), 0, index + 1);
    }
}

 

LeetCode 411: Minimum Unique Word Abbreviation

标签:one   return   bsp   rate   span   val   word   class   valueof   

原文地址:http://www.cnblogs.com/shuashuashua/p/7662213.html

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