Top K Frequent Elements 347. Top K Frequent Elements [LeetCode] Top K Frequent Elements 前K个高频元素…
Given a non-empty array of integers, return the k most frequent elements. For example,Given [1,1,1,2,2,3] and k = 2, return [1,2]. Note: You may assume k is always valid, 1 ≤ k ≤ number of unique elements.    Your algorithm's time complexity must be…
Given a non-empty list of words, return the k most frequent elements. Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first. Example 1: Inpu…
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 ele…
给定一个非空的整数数组,返回其中出现频率前 k 高的元素.例如,给定数组 [1,1,1,2,2,3] , 和 k = 2,返回 [1,2].注意:    你可以假设给定的 k 总是合理的,1 ≤ k ≤ 数组中不相同的元素的个数.    你的算法的时间复杂度必须优于 O(n log n) , n 是数组的大小.详见:https://leetcode.com/problems/top-k-frequent-elements/description/C++: class Solution { pub…
这个题的排序是用的PriorityQueue实现自动排列,优先队列用的是堆排序,堆排序请看:http://www.cnblogs.com/stAr-1/p/7569706.html 自定义了优先队列的排序方法,lambda表达式该复习下了... public List<String> topKFrequent(String[] words, int k) { /* 1.建立字符串和出现频率的映射 2.用优先队列PriorityQueue实现排序(相同频率用compareTo方法) 4.挑选出前…
Given a non-empty list of words, return the k most frequent elements. Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first. Example 1: Inpu…
https://leetcode.com/problems/top-k-frequent-words/ Given a non-empty list of words, return the k most frequent elements. Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the l…
基础知识 ArrayDeque deque = new ArrayDeque(); /* offerFirst(E e) 在数组前面添加元素,并返回是否添加成功 offerLast(E e) 在数组后天添加元素,并返回是否添加成功 pollFirst()删除第一个元素,并返回删除元素的值,如果元素为null,将返回null pollLast()删除最后一个元素,并返回删除元素的值,如果为null,将返回null getFirst() 获取第一个元素,如果没有将抛出异常 getLast() 获取最…
题目:设计一组N个数,确定其中第k个最大值 1,普通方法(先排序,然后遍历,得到第k大的数)      注:如果是数组,直接arr[k],我们可以对这个乱序数组按照从大到小先行排序,然后取出前k大,总的时间复杂度为O(n*logn + k),即O(n*logn ) 2.利用部分排序,以避免N-K个数字的排序,例如用选择排序,或者冒泡排序,K次选择后即可得到第k大的数.总的时间复杂度为O(n*k) 此方法在k很小的时候,效率不错 3.利用堆排序:建立大顶堆,每次弹出最大值,然后调整堆,再次建立大顶…