LeetCode-Python-1268. 搜索推荐系统( 二分查找 + 字符串)

 

给你一个产品数组 products 和一个字符串 searchWord ,products  数组中每个产品都是一个字符串。

请你设计一个推荐系统,在依次输入单词 searchWord 的每一个字母后,推荐 products 数组中前缀与 searchWord 相同的最多三个产品。如果前缀相同的可推荐产品超过三个,请按字典序返回最小的三个。

请你以二维列表的形式,返回在输入 searchWord 每个字母后相应的推荐产品的列表。

 

示例 1:

输入:products = ["mobile","mouse","moneypot","monitor","mousepad"], searchWord = "mouse"
输出:[
["mobile","moneypot","monitor"],
["mobile","moneypot","monitor"],
["mouse","mousepad"],
["mouse","mousepad"],
["mouse","mousepad"]
]
解释:按字典序排序后的产品列表是 ["mobile","moneypot","monitor","mouse","mousepad"]
输入 m 和 mo,由于所有产品的前缀都相同,所以系统返回字典序最小的三个产品 ["mobile","moneypot","monitor"]
输入 mou, mous 和 mouse 后系统都返回 ["mouse","mousepad"]
示例 2:

输入:products = ["havana"], searchWord = "havana"
输出:[["havana"],["havana"],["havana"],["havana"],["havana"],["havana"]]
示例 3:

输入:products = ["bags","baggage","banner","box","cloths"], searchWord = "bags"
输出:[["baggage","bags","banner"],["baggage","bags","banner"],["baggage","bags"],["bags"]]
示例 4:

输入:products = ["havana"], searchWord = "tatiana"
输出:[[],[],[],[],[],[],[]]
 

提示:

1 <= products.length <= 1000
1 <= Σ products[i].length <= 2 * 10^4
products[i] 中所有的字符都是小写英文字母。
1 <= searchWord.length <= 1000
searchWord 中所有字符都是小写英文字母。

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/search-suggestions-system
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

思路:

先一劳永逸把products排好序,

然后线性扫描searchWord, 不断累加prefix,

再在products里利用二分找前缀是prefix单词的起点,然后往后找最多三个相同前缀的单词。

时间复杂度:O(MlogM + N * logM)

空间复杂度:O(N + 3 * max(len(item) for item in products))

class Solution(object):
    def suggestedProducts(self, products, searchWord):
        """
        :type products: List[str]
        :type searchWord: str
        :rtype: List[List[str]]
        """
        products.sort()
        res = []
        prefix = ""
        for char in searchWord:
            tmp = []
            prefix += char
            idx = bisect.bisect_left(products, prefix) # 找当前prefix 起点,因为有序所以二分
            for word in products[idx:]:
                if len(tmp) >= 3 or word[:len(prefix)] > prefix:
                    break
                if word[:len(prefix)] == prefix:
                    tmp.append(word)
            res.append(tmp)
        return res

 


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