给定一个字符串数组,再给定整数 k ,请返回出现次数前k名的字符串和对应的次数。
返回的答案应该按字符串出现频率由高到低排序。如果不同的字符串有相同出现频率,按字典序排序。
对于两个字符串,大小关系取决于两个字符串从左到右第一个不同字符的 ASCII 值的大小关系。
比如"ah1x"小于"ahb","231"<”32“
字符仅包含数字和字母
数据范围:字符串数满足 1000000≤n≤100000,每个字符串长度 100≤n≤10, 25000≤k≤2500
要求:空间复杂度 O(n),时间复杂度O(nlogk)
输入:
["a","b","c","b"],2
返回值:
[["b","2"],["a","1"]]
说明:
"b"出现了2次,记["b","2"],"a"与"c"各出现1次,但是a字典序在c前面,记["a","1"],最后返回[["b","2"],["a","1"]]
输入:
["123","123","231","32"],2
返回值:
[["123","2"],["231","1"]]
说明:
"123"出现了2次,记["123","2"],"231"与"32"各出现1次,但是"231"字典序在"32"前面,记["231","1"],最后返回[["123","2"],["231","1"]]
输入:
["abcd","abcd","abcd","pwb2","abcd","pwb2","p12"],3
返回值:
[["abcd","4"],["pwb2","2"],["p12","1"]]
1≤N≤105
Python代码:
#
# 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
#
# return topK string
# @param strings string字符串一维数组 strings
# @param k int整型 the k
# @return string字符串二维数组
#
from collections import Counter
class Solution:
def topKstrings(self , strings: List[str], k: int) -> List[List[str]]:
# write code here
counter = Counter(strings)
return sorted(counter.items(), key=lambda x: (-x[1], x[0]))[:k]