leecode 解题总结：347. Top K Frequent Elements

#include <iostream>#include <stdio.h>#include <vector>#include <string>#include <unordered_map>using namespace std;/*问题: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 better than O(n log n), where n is the array's size.分析:此题要求计算数组中前k个出现次数的元素。时间复杂度要好于O(n log n)。如果用统计排序，那么由于没有告诉数组中元素值的范围，并不好计算，会溢出。至少需要记录每个元素出现的次数，然后选择前K个。用map。map本质上是红黑树，插入的时间复杂度为O(logN)，查找的时间复杂度为O(logN)。遍历数组，插入map,时间复杂度为O(N*logN)。遍历map，寻找前k个出现次数最多的，时间复杂度O(N*logN)此题和寻找前k个最大的数不同，寻找前k个最大的数用优先级队列(本质是堆)输入:6(数组元素个数) 2(k)1 1 1 2 2 3输出:1 2报错:Input:[1,2]2Output:[1]Expected:[1,2]关键1 参考:https://leetcode.com/problems/top-k-frequent-elements/?tab=Solutions采用哈希建立<数字，出现次数>的映射，然后出现次数最多不会超过n,建立一个数组数组中A[i]表示出现次数为i的元素为A[i]，将该数组逆序输出k个即可。哈希统计，O(n),存入数组O(n)。厉害，建立了两次映射。  实际是对次数建立桶排序*/class Solution {public:    vector<int> topKFrequent(vector<int>& nums, int k) {vector<int> result;if(nums.empty() || k <= 0){return result;}        unordered_map<int , int> numToTimes;int size = nums.size();//统计数字到出现次数的映射for(int i = 0 ; i < size ; i++){if(numToTimes.find(nums.at(i)) != numToTimes.end()){numToTimes[nums.at(i)]++;}else{numToTimes[nums.at(i)] = 1;}}//建立<出现次数，[数字]>的映射，从后向前遍历k个即为所求,但是出现次数可能是相同的，数字必须用数组表示vector< vector<int> > timesToNum(size + 1 , vector<int>());for(unordered_map<int , int>::iterator it = numToTimes.begin(); it != numToTimes.end() ; it++){timesToNum.at(it->second).push_back(it->first);}int count = 0;for(int i = size ; i >= 0 ; i--){if(!timesToNum.at(i).empty() ){int len = timesToNum.at(i).size();for(int j = 0 ; j < len ; j++){if(count < k){result.push_back( timesToNum.at(i).at(j)  );count++;}if(count >= k){return result;}}}}return result;    }};void print(vector<int>& result){if(result.empty()){cout << "no result" << endl;return;}int size = result.size();for(int i = 0 ; i < size ; i++){cout << result.at(i) << " " ;}cout << endl;}void process(){ vector<int> nums; int value; int num; Solution solution; vector<int> result; int k; while(cin >> num >> k ) { nums.clear(); for(int i = 0 ; i < num ; i++) { cin >> value; nums.push_back(value); } result = solution.topKFrequent(nums , k); print(result); }}int main(int argc , char* argv[]){process();getchar();return 0;}