1101. Quick Sort (25)
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1.这道题目需要考虑采用适当的数据结构,即小根堆和大根堆
2.判断某个元素是否能够成为pivot,那么该元素左边数组应该构成一个大根堆,堆顶元素应该小于该元素,该元素的右边构成小根堆,堆顶元素大于该元素
3.左边的小根堆,一直插入即可,利用priority_queue,而右边则需要进行维护,需要编写仿函数
4右边的大根堆维护机制:建立哈希表times,记录每个元素出现的次数,每往后检测新元素数,把新元素出现的次数-1,同时检测大根堆的堆顶元素,如果出现次数为0则弹出,直至剩下出现次数不为0的堆顶
AC代码如下:
//#include<string>//#include <iomanip>#include<vector>#include <algorithm>//#include<stack>#include<set>#include<queue>#include<map>//#include<unordered_set>#include<unordered_map>//#include <sstream>//#include "func.h"//#include <list>#include<stdio.h>#include<iostream>#include<string>#include<memory.h>#include<limits.h>using namespace std;struct cmp{bool operator()(const int&a, const int&b){return a > b;}};int main(void){int sum;cin >> sum;int *num = new int[sum];map<int, int> times;/*vector<int> times(100001, 0);*/priority_queue<int> lq;priority_queue<int,vector<int>,cmp> rq;for (int i = 0; i < sum; i++){scanf("%d", &num[i]);times[num[i]]++;rq.push(num[i]);}vector<int> ans(0);for (int i = 0; i < sum; i++){if (i>0) lq.push(num[i - 1]);if (rq.size() != 0){//如果右边heap不为空times[num[i]]--;//减少num[i]的次数,以维护右边的小根堆while (rq.size() != 0 && times[rq.top()] == 0) rq.pop();//检测堆顶元素,出现次数是否为0,如果为0,证明应该被弹出,if (rq.size() != 0 && num[i]>rq.top()){continue;}}if (lq.size() != 0 && num[i] < lq.top()){continue;}ans.push_back(num[i]);}cout << ans.size() << endl;sort(ans.begin(), ans.end());for (int i = 0; i < ans.size(); i++){printf("%d", ans[i]);if (i != ans.size() - 1)cout << " ";}cout << endl;//注意在后面添加换行return 0;}
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- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101. Quick Sort (25)
- 1101.Quick Sort (25)
- 1101. Quick Sort (25)
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