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添加 problem 347
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package leetcode
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import "container/heap"
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func topKFrequent(nums []int, k int) []int {
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m := make(map[int]int)
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for _, n := range nums {
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m[n]++
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}
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q := PriorityQueue{}
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for key, count := range m {
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heap.Push(&q, &Item{key: key, count: count})
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}
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var result []int
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for len(result) < k {
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item := heap.Pop(&q).(*Item)
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result = append(result, item.key)
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}
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return result
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}
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type Item struct {
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key int
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count int
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}
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// A PriorityQueue implements heap.Interface and holds Items.
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type PriorityQueue []*Item
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func (pq PriorityQueue) Len() int {
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return len(pq)
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}
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func (pq PriorityQueue) Less(i, j int) bool {
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// 注意:因为golang中的heap是按最小堆组织的,所以count越大,Less()越小,越靠近堆顶.
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return pq[i].count > pq[j].count
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}
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func (pq PriorityQueue) Swap(i, j int) {
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pq[i], pq[j] = pq[j], pq[i]
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}
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func (pq *PriorityQueue) Push(x interface{}) {
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item := x.(*Item)
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*pq = append(*pq, item)
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}
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func (pq *PriorityQueue) Pop() interface{} {
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n := len(*pq)
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item := (*pq)[n-1]
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*pq = (*pq)[:n-1]
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return item
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}
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@@ -0,0 +1,48 @@
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package leetcode
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import (
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"fmt"
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"testing"
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)
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type question347 struct {
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para347
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ans347
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}
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// para 是参数
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// one 代表第一个参数
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type para347 struct {
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one []int
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two int
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}
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// ans 是答案
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// one 代表第一个答案
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type ans347 struct {
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one []int
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}
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func Test_Problem347(t *testing.T) {
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qs := []question347{
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question347{
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para347{[]int{1, 1, 1, 2, 2, 3}, 2},
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ans347{[]int{1, 2}},
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},
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question347{
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para347{[]int{1}, 1},
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ans347{[]int{1}},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 347------------------------\n")
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for _, q := range qs {
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_, p := q.ans347, q.para347
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fmt.Printf("【input】:%v 【output】:%v\n", p, topKFrequent(p.one, p.two))
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}
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fmt.Printf("\n\n\n")
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}
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33
Algorithms/347. Top K Frequent Elements/README.md
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33
Algorithms/347. Top K Frequent Elements/README.md
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# [347. Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements/)
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## 题目
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Given a non-empty array of integers, return the k most frequent elements.
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Example 1:
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```c
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Input: nums = [1,1,1,2,2,3], k = 2
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Output: [1,2]
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```
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Example 2:
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```c
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Input: nums = [1], k = 1
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Output: [1]
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```
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Note:
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- You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
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- Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
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## 题目大意
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给一个非空的数组,输出前 K 个频率最高的元素。
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这一题是考察优先队列的题目。把数组构造成一个优先队列,输出前 K 个即可。
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