comments | difficulty | edit_url | tags | |||||
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中等 |
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请设计一个类,使该类的构造函数能够接收一个字符串数组。然后再实现一个方法,该方法能够分别接收两个单词,并返回列表中这两个单词之间的最短距离。
实现 WordDistanc
类:
WordDistance(String[] wordsDict)
用字符串数组wordsDict
初始化对象。int shortest(String word1, String word2)
返回数组worddict
中word1
和word2
之间的最短距离。
示例 1:
输入: ["WordDistance", "shortest", "shortest"] [[["practice", "makes", "perfect", "coding", "makes"]], ["coding", "practice"], ["makes", "coding"]] 输出: [null, 3, 1] 解释: WordDistance wordDistance = new WordDistance(["practice", "makes", "perfect", "coding", "makes"]); wordDistance.shortest("coding", "practice"); // 返回 3 wordDistance.shortest("makes", "coding"); // 返回 1
注意:
1 <= wordsDict.length <= 3 * 104
1 <= wordsDict[i].length <= 10
wordsDict[i]
由小写英文字母组成word1
和word2
在数组wordsDict
中word1 != word2
-
shortest
操作次数不大于5000
我们用哈希表
初始化的时间复杂度为 shortest
方法的时间复杂度为
class WordDistance:
def __init__(self, wordsDict: List[str]):
self.d = defaultdict(list)
for i, w in enumerate(wordsDict):
self.d[w].append(i)
def shortest(self, word1: str, word2: str) -> int:
a, b = self.d[word1], self.d[word2]
ans = inf
i = j = 0
while i < len(a) and j < len(b):
ans = min(ans, abs(a[i] - b[j]))
if a[i] <= b[j]:
i += 1
else:
j += 1
return ans
# Your WordDistance object will be instantiated and called as such:
# obj = WordDistance(wordsDict)
# param_1 = obj.shortest(word1,word2)
class WordDistance {
private Map<String, List<Integer>> d = new HashMap<>();
public WordDistance(String[] wordsDict) {
for (int i = 0; i < wordsDict.length; ++i) {
d.computeIfAbsent(wordsDict[i], k -> new ArrayList<>()).add(i);
}
}
public int shortest(String word1, String word2) {
List<Integer> a = d.get(word1), b = d.get(word2);
int ans = 0x3f3f3f3f;
int i = 0, j = 0;
while (i < a.size() && j < b.size()) {
ans = Math.min(ans, Math.abs(a.get(i) - b.get(j)));
if (a.get(i) <= b.get(j)) {
++i;
} else {
++j;
}
}
return ans;
}
}
/**
* Your WordDistance object will be instantiated and called as such:
* WordDistance obj = new WordDistance(wordsDict);
* int param_1 = obj.shortest(word1,word2);
*/
class WordDistance {
public:
WordDistance(vector<string>& wordsDict) {
for (int i = 0; i < wordsDict.size(); ++i) {
d[wordsDict[i]].push_back(i);
}
}
int shortest(string word1, string word2) {
auto a = d[word1], b = d[word2];
int i = 0, j = 0;
int ans = INT_MAX;
while (i < a.size() && j < b.size()) {
ans = min(ans, abs(a[i] - b[j]));
if (a[i] <= b[j]) {
++i;
} else {
++j;
}
}
return ans;
}
private:
unordered_map<string, vector<int>> d;
};
/**
* Your WordDistance object will be instantiated and called as such:
* WordDistance* obj = new WordDistance(wordsDict);
* int param_1 = obj->shortest(word1,word2);
*/
type WordDistance struct {
d map[string][]int
}
func Constructor(wordsDict []string) WordDistance {
d := map[string][]int{}
for i, w := range wordsDict {
d[w] = append(d[w], i)
}
return WordDistance{d}
}
func (this *WordDistance) Shortest(word1 string, word2 string) int {
a, b := this.d[word1], this.d[word2]
ans := 0x3f3f3f3f
i, j := 0, 0
for i < len(a) && j < len(b) {
ans = min(ans, abs(a[i]-b[j]))
if a[i] <= b[j] {
i++
} else {
j++
}
}
return ans
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
/**
* Your WordDistance object will be instantiated and called as such:
* obj := Constructor(wordsDict);
* param_1 := obj.Shortest(word1,word2);
*/