692. Top K Frequent Words

Medium

Given an array of strings words and an integer k, return the k most frequent strings.

Return the answer sorted by the frequency from highest to lowest. Sort the words with the same frequency by their lexicographical order.

Example 1:

Input: words = ["i","love","leetcode","i","love","coding"], k = 2
Output:
 ["i","love"]
Explanation:
 "i" and "love" are the two most frequent words.
Note that "i" comes before "love" due to a lower alphabetical order.

Example 2:

Input: words = ["the","day","is","sunny","the","the","the","sunny","is","is"], k = 4
Output:
 ["the","is","sunny","day"]
Explanation:
 "the", "is", "sunny" and "day" are the four most frequent words, with the number of occurrence being 4, 3, 2 and 1 respectively.

Constraints:

• 1 <= words.length <= 500

• 1 <= words[i].length <= 10

• words[i] consists of lowercase English letters.

• k is in the range [1, The number of unique words[i]]

Follow-up: Could you solve it in O(n log(k)) time and O(n) extra space?

解題

func topKFrequent(words []string, k int) []string {
    count := make(map[string]int)
    
    for _, val := range words {
        if _,ok := count[val]; ok {
            count[val]++
        } else {
            count[val] = 1
        }
    }

    type kv struct {
        Key   string
        Value int
    }

    var ss []kv
    for k, v := range count {
        ss = append(ss, kv{k, v})
    }

    sort.Slice(ss, func(i, j int) bool {
        if ss[i].Value==ss[j].Value {
            return ss[i].Key < ss[j].Key
        }
        return ss[i].Value > ss[j].Value
    })

    ans := []string{}
    for i:=0; i<k; i++ {
        ans = append(ans, ss[i].Key)
    }
    
    return ans
}

也可以不用struct。

func topKFrequent(words []string, k int) []string {
    count := make(map[string]int)
    
    for _, val := range words {
        if _,ok := count[val]; ok {
            count[val]++
        } else {
            count[val] = 1
        }
    }

    keys := make([]string, 0, len(count))
    for key := range count {
        keys = append(keys, key)
    }
    sort.Slice(keys, func(i, j int) bool { 
        if count[keys[i]] == count[keys[j]] {
            return keys[i] < keys[j]
        }
        return count[keys[i]] > count[keys[j]] 
    })
    
    return keys[:k]
}