LintCode 471 [Top K Frequent Words]

原题

Given a list of words and an integer k, return the top k frequent words in the list.

给出

[ 
"yes", "lint", "code", 
"yes", "code", "baby", 
"you", "baby", "chrome", 
"safari", "lint", "code", 
"body", "lint", "code"
]

当k=3时，返回["code", "lint", "baby"]
当k=4时，返回["code", "lint", "baby", "yes"]

解题思路

• 首先使用hash table统计词频，建立一个存有（单词，词频）结构的数组
• 自定义（单词，词频）结构的比较方式
• 快速排序，然后取出前K个，得到结果

完整代码

class Solution:
    # @param {string[]} words a list of string
    # @param {int} k an integer
    # @return {string[]} a list of string
    def topKFrequentWords(self, words, k):
        map = {}
        for word in words:
            if word not in map:
                map[word] = 1
            else:
                map[word] += 1
        temp = []
        for key, value in map.items():
            temp.append((value, key))
        temp.sort(cmp=self.cmp)
        result = []
        for i in range(k):
            result.append(temp[i][1])
        return result
        
    def cmp(self, a, b):
        if a[0] > b[0] or a[0] == b[0] and a[1] < b[1]:
            return -1
        elif a[0] == b[0] and a[1] == b[1]:
            return 0
        else:
            return 1