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