Question

244. 最短单词距离 II

English Version

题目描述

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 

解法

方法一：哈希表 + 双指针

我们用哈希表 $d$ 存储每个单词在数组中出现的所有下标，然后用双指针 $i$ 和 $j$ 分别指向两个单词在数组中出现的下标列表 $a$ 和 $b$，每次更新下标差值的最小值，然后移动下标较小的指针，直到其中一个指针遍历完下标列表。

初始化的时间复杂度为 $O(n)$，其中 $n$ 为数组的长度。每次调用 shortest 方法的时间复杂度为 $O(m + n)$，其中 $m$ 为两个单词在数组中出现的下标列表的长度之和。

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);
 */