|
| 1 | +/** |
| 2 | + Tries 자료구조를 구현하고, 이를 통해 단어를 탐색하는 방식 |
| 3 | + visited[] 방문 배열을 추가하여, 이전에 방문했던 인덱스에 대해서는 무시 |
| 4 | + */ |
| 5 | +class WordMap { |
| 6 | + public Map<Character, WordNode> wordMap; |
| 7 | + |
| 8 | + public WordMap() { |
| 9 | + wordMap = new HashMap<>(); |
| 10 | + } |
| 11 | + |
| 12 | + public List<Integer> search(String word, int idx) { |
| 13 | + List<Integer> idxList = new ArrayList<>(); |
| 14 | + |
| 15 | + WordNode wordNode = null; |
| 16 | + char ch = word.charAt(idx); |
| 17 | + wordNode = wordMap.get(ch); |
| 18 | + if (wordNode == null) return idxList; |
| 19 | + |
| 20 | + if(wordNode.isLeaf) { |
| 21 | + idxList.add(idx + 1); |
| 22 | + } |
| 23 | + idx++; |
| 24 | + |
| 25 | + for(; idx < word.length(); idx++) { |
| 26 | + char target = word.charAt(idx); |
| 27 | + if (!wordNode.next.containsKey(target)) { |
| 28 | + break; |
| 29 | + } |
| 30 | + wordNode = wordNode.next.get(target); |
| 31 | + if (wordNode.isLeaf) { |
| 32 | + idxList.add(idx + 1); |
| 33 | + } |
| 34 | + } |
| 35 | + |
| 36 | + return idxList; |
| 37 | + } |
| 38 | + |
| 39 | + public void add(String word) { |
| 40 | + WordNode wordNode = null; |
| 41 | + char ch = word.charAt(0); |
| 42 | + wordNode = wordMap.get(ch); |
| 43 | + |
| 44 | + if(wordNode == null) { |
| 45 | + boolean isFirstWord = word.length() == 1; |
| 46 | + wordNode = new WordNode(ch, isFirstWord); |
| 47 | + wordMap.put(ch, wordNode); |
| 48 | + } |
| 49 | + |
| 50 | + for(int idx = 1; idx < word.length(); idx++) { |
| 51 | + char target = word.charAt(idx); |
| 52 | + boolean isLeaf = word.length() - 1 == idx; |
| 53 | + wordNode = wordNode.next.computeIfAbsent(target, key -> new WordNode(key, isLeaf)); |
| 54 | + } |
| 55 | + |
| 56 | + wordNode.isLeaf = true; |
| 57 | + |
| 58 | + } |
| 59 | + |
| 60 | +} |
| 61 | + |
| 62 | +class WordNode { |
| 63 | + char ch; |
| 64 | + Map<Character, WordNode> next; |
| 65 | + boolean isLeaf; |
| 66 | + |
| 67 | + public WordNode(char ch) { |
| 68 | + this(ch, false); |
| 69 | + } |
| 70 | + |
| 71 | + public WordNode(char ch, boolean isLeaf) { |
| 72 | + next = new HashMap<>(); |
| 73 | + this.ch = ch; |
| 74 | + this.isLeaf = isLeaf; |
| 75 | + } |
| 76 | + |
| 77 | +} |
| 78 | + |
| 79 | +class Solution { |
| 80 | + public static WordMap wordMap; |
| 81 | + public boolean wordBreak(String s, List<String> wordDict) { |
| 82 | + boolean[] visited = new boolean[s.length()]; |
| 83 | + initWordMap(wordDict); |
| 84 | + if(s.length() == 1) return wordMap.wordMap.containsKey(s.charAt(0)) && wordMap.wordMap.get(s.charAt(0)).isLeaf; |
| 85 | + |
| 86 | + Queue<Integer> que = new LinkedList<>(); |
| 87 | + boolean result = false; |
| 88 | + que.add(0); |
| 89 | + visited[0] = true; |
| 90 | + loop: |
| 91 | + while(!que.isEmpty()) { |
| 92 | + int idx = que.poll(); |
| 93 | + List<Integer> idxList = wordMap.search(s, idx); |
| 94 | + for(int i : idxList) { |
| 95 | + if(i == s.length()) { |
| 96 | + result = true; |
| 97 | + break loop; |
| 98 | + } |
| 99 | + |
| 100 | + if(!visited[i]) { |
| 101 | + que.add(i); |
| 102 | + visited[i] = true; |
| 103 | + } |
| 104 | + } |
| 105 | + } |
| 106 | + |
| 107 | + return result; |
| 108 | + |
| 109 | + |
| 110 | + |
| 111 | + } |
| 112 | + |
| 113 | + public void initWordMap(List<String> wordDict) { |
| 114 | + wordMap = new WordMap(); |
| 115 | + for(String word : wordDict) { |
| 116 | + wordMap.add(word); |
| 117 | + } |
| 118 | + } |
| 119 | +} |
| 120 | + |
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