Given a non-empty array of integers, return the k most frequent elements.
Example 1:
Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]
Example 2:
Input: nums = [1], k = 1
Output: [1]
Note:
- You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
- Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
- It's guaranteed that the answer is unique, in other words the set of the top k frequent elements is unique.
- You can return the answer in any order.
1
2
3
4
5
6
7
8
9
|
from collections import Counter
class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
if k == len(nums):
return nums
count = Counter(nums)
return heapq.nlargest(k, count.keys(), key=count.get)
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
|
func topKFrequent(nums []int, k int) []int {
// count frequencies
m := make(map[int]int)
for _, n := range nums {
m[n] += 1
}
// build a min-heap
var heap Heap
for num, count := range m {
heap.pushWithMax(num, count, k)
}
// extract heap values
result := make([]int, len(heap))
for i, node := range heap {
result[i] = node.Num
}
return result
}
type Node struct {
Num int
Count int
}
// min-heap
type Heap []*Node
func (heap *Heap) pushWithMax(num, count, maxSize int) {
heap.push(&Node{
Num: num,
Count: count,
})
if len(*heap) > maxSize {
heap.popMin()
}
}
func (heap *Heap) push(node *Node) {
*heap = append(*heap, node)
i := len(*heap) - 1
// shift up
for i > 0 {
parent := (i - 1) / 2
if (*heap)[parent].Count > (*heap)[i].Count {
heap.swap(parent, i)
i = parent
} else {
break
}
}
}
func (heap *Heap) popMin() {
h := *heap
// move last to top
(h)[0] = h[len(h)-1]
// drop last
*heap = h[:len(h)-1]
// shift down
totalLen := len(h)
i := 0
var smallest int
for i < totalLen {
li := i*2 + 1
ri := i*2 + 2
smallest = i
if li < totalLen && h[li].Count < h[i].Count {
smallest = li
}
if ri < totalLen && h[ri].Count < h[smallest].Count {
smallest = ri
}
if smallest != i {
heap.swap(smallest, i)
i = smallest
continue
}
break
}
}
func (heap *Heap) swap(i, j int) {
(*heap)[i], (*heap)[j] = (*heap)[j], (*heap)[i]
}
|