745. 前綴和後綴搜索
給定多個 words,words[i] 的權重爲 i 。
設計一個類 WordFilter 實現函數WordFilter.f(String prefix, String suffix)。這個函數將返回具有前綴 prefix 和後綴suffix 的詞的最大權重。如果沒有這樣的詞,返回 -1。
例子:
輸入:
WordFilter([“apple”])
WordFilter.f(“a”, “e”) // 返回 0
WordFilter.f(“b”, “”) // 返回 -1
注意:
words的長度在[1, 15000]之間。
對於每個測試用例,最多會有words.length次對WordFilter.f的調用。
words[i]的長度在[1, 10]之間。
prefix, suffix的長度在[0, 10]之前。
words[i]和prefix, suffix只包含小寫字母。
class WordFilter {
HashMap<String, List<Integer>> prefMap = new HashMap<>();
HashMap<String, List<Integer>> suffMap = new HashMap<>();
String[] words;
void addToPref(String word, int idx) {
int wlen = word.length();
for (int i = 1; i <= wlen; i++) {
prefMap.computeIfAbsent(word.substring(0, i), k -> new ArrayList<>()).add(idx);
}
}
void addToSuff(String word, int idx) {
int wlen = word.length();
for (int i = 0; i < wlen; i++) {
suffMap.computeIfAbsent(word.substring(i), k -> new ArrayList<>()).add(idx);
}
}
public WordFilter(String[] words) {
int size = words.length;
this.words = words;
for (int i = 0; i < size; i++) {
addToPref(words[i], i);
addToSuff(words[i], i);
}
}
public int f(String prefix, String suffix) {
List<Integer> l1 = prefMap.get(prefix);
List<Integer> l2 = suffMap.get(suffix);
if (prefix.length() == 0 || suffix.length() == 0) {
if (prefix.length() == 0 && suffix.length() == 0) {
return words.length-1;
}
if (prefix.length() == 0) {
return l2 == null ? -1 : l2.get(l2.size()-1);
}
return l1 == null ? -1 : l1.get(l1.size()-1);
}
if (l1 == null || l2 == null) return -1;
int idx1 = l1.size()-1;
int idx2 = l2.size()-1;
while (idx1 >= 0 && idx2 >= 0) {
int i1 = l1.get(idx1);
int i2 = l2.get(idx2);
if (i1 == i2) return i1;
if (i1 < i2)
idx2--;
else idx1--;
}
return -1;
}
}
/**
* Your WordFilter object will be instantiated and called as such:
* WordFilter obj = new WordFilter(words);
* int param_1 = obj.f(prefix,suffix);
*/