Open Shawngbk opened 7 years ago
因为在一个数组里可能会有重复的元素，所以他的Key值是一组数据，是一个ArrayList public class WordDistance { Map<String,List> map = new HashMap<String, List>(); public WordDistance(String[] words) { for(int i = 0; i < words.length; i++) { List temp = map.get(words[i]); if(temp == null) { temp = new ArrayList(); } temp.add(i); map.put(words[i], temp); } }
public int shortest(String word1, String word2) { List<Integer> index1 = map.get(word1); List<Integer> index2 = map.get(word2); int distance = Integer.MAX_VALUE; int i = 0, j = 0; while(i < index1.size() && j < index2.size()) { distance = Math.min(distance, Math.abs(index1.get(i)-index2.get(j))); if(index1.get(i) < index2.get(j)) { i++; } else { j++; } } return distance; }
}
// Your WordDistance object will be instantiated and called as such: // WordDistance wordDistance = new WordDistance(words); // wordDistance.shortest("word1", "word2"); // wordDistance.shortest("anotherWord1", "anotherWord2");
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Shawngbk
commented
7 years ago
因为在一个数组里可能会有重复的元素，所以他的Key值是一组数据，是一个ArrayList public class WordDistance { Map<String,List> map = new HashMap<String, List>();
public WordDistance(String[] words) {
for(int i = 0; i < words.length; i++) {
List temp = map.get(words[i]);
if(temp == null) {
temp = new ArrayList();
}
temp.add(i);
map.put(words[i], temp);
}
}
}
// Your WordDistance object will be instantiated and called as such: // WordDistance wordDistance = new WordDistance(words); // wordDistance.shortest("word1", "word2"); // wordDistance.shortest("anotherWord1", "anotherWord2");