244. 最短单词距离 II 🔒

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。

Python3JavaC++Go

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);
 */

244. 最短单词距离 II 🔒

题目描述

解法

方法一：哈希表 + 双指针

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