commit 4

Author: tkzzzzzz6

book/代码随想录/栈和队列/347.前 K 个高频元素/1.md

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+<!--
+ * @Author: tkzzzzzz6
+ * @Date: 2026-04-19 14:25:47
+ * @LastEditors: tkzzzzzz6
+ * @LastEditTime: 2026-04-19 14:25:50
+-->
+
+大小顶堆:二叉树
+
+优先队列
+
+不用维护所有元素,只需要维持 k 个元素
+
+这里推荐使用小顶堆,因为我们需要维护 k 个最大的元素,小顶堆的堆顶就是最小的元素,当我们遇到一个新的元素时,如果它比堆顶元素大,说明它应该进入堆中,我们就把堆顶元素弹出,然后把新的元素加入堆中。这样我们就能保证堆中始终保持 k 个最大的元素。

book/代码随想录/栈和队列/347.前 K 个高频元素/347.top-k-frequent-elements.cpp

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+/*
+ * @lc app=leetcode.cn id=347 lang=cpp
+ * @lcpr version=30204
+ *
+ * [347] 前 K 个高频元素
+ */
+
+
+// @lcpr-template-start
+using namespace std;
+#include <algorithm>
+#include <array>
+#include <bitset>
+#include <climits>
+#include <deque>
+#include <functional>
+#include <iostream>
+#include <list>
+#include <queue>
+#include <stack>
+#include <tuple>
+#include <unordered_map>
+#include <unordered_set>
+#include <utility>
+#include <vector>
+// @lcpr-template-end
+// @lc code=start
+class Solution {
+public:
+    vector<int> topKFrequent(vector<int>& nums, int k) {
+        unordered_map<int,int> mp;
+        int n = nums.size();
+
+        for(int i = 0;i<n;++i)
+            mp[nums[i]]++;
+
+        auto cmp = [](const auto&a,const auto&b){
+            return a.second > b.second;
+        };
+
+        priority_queue<pair<int,int>,vector<pair<int,int>>,decltype(cmp)> pq(cmp);
+
+        for(unordered_map<int,int>::iterator it = mp.begin();it != mp.end();++it){
+            pq.push(*it);
+            if(pq.size() > k){
+                pq.pop();
+            }
+        }
+
+        vector<int> res(k);
+        for(int i = k-1;i>=0;--i){
+            res[i] = pq.top().first;
+            pq.pop();
+        }
+        return res;
+    }
+};
+// @lc code=end
+
+
+
+/*
+// @lcpr case=start
+// [1,1,1,2,2,3]\n2\n
+// @lcpr case=end
+
+// @lcpr case=start
+// [1]\n1\n
+// @lcpr case=end
+
+// @lcpr case=start
+// [1,2,1,2,1,2,3,1,3,2]\n2\n
+// @lcpr case=end
+
+ */

leetcode/problems/.lcpr_data/bricks.json

@@ -0,0 +1,19 @@
+{
+    "version": 1,
+    "all_bricks": {
+        "347": {
+            "submit_time": [
+                1776593902
+            ],
+            "type": 4,
+            "review_day": [
+                1776614400,
+                1776873600,
+                1777132800,
+                1777737600,
+                1778947200,
+                1781712000
+            ]
+        }
+    }
+}

leetcode/problems/.lcpr_data/group.json

@@ -0,0 +1 @@
+{"version":1}

leetcode/problems/347.top-k-frequent-elements.cpp

@@ -0,0 +1,75 @@
+/*
+ * @lc app=leetcode.cn id=347 lang=cpp
+ * @lcpr version=30204
+ *
+ * [347] 前 K 个高频元素
+ */
+
+
+// @lcpr-template-start
+using namespace std;
+#include <algorithm>
+#include <array>
+#include <bitset>
+#include <climits>
+#include <deque>
+#include <functional>
+#include <iostream>
+#include <list>
+#include <queue>
+#include <stack>
+#include <tuple>
+#include <unordered_map>
+#include <unordered_set>
+#include <utility>
+#include <vector>
+// @lcpr-template-end
+// @lc code=start
+class Solution {
+public:
+    vector<int> topKFrequent(vector<int>& nums, int k) {
+        unordered_map<int,int> mp;
+        int n = nums.size();
+
+        for(int i = 0;i<n;++i)
+            mp[nums[i]]++;
+
+        auto cmp = [](const auto&a,const auto&b){
+            return a.second > b.second;
+        };
+
+        priority_queue<pair<int,int>,vector<pair<int,int>>,decltype(cmp)> pq(cmp);
+
+        for(unordered_map<int,int>::iterator it = mp.begin();it != mp.end();++it){
+            pq.push(*it);
+            if(pq.size() > k){
+                pq.pop();
+            }
+        }
+
+        vector<int> res(k);
+        for(int i = k-1;i>=0;--i){
+            res[i] = pq.top().first;
+            pq.pop();
+        }
+        return res;
+    }
+};
+// @lc code=end
+
+
+
+/*
+// @lcpr case=start
+// [1,1,1,2,2,3]\n2\n
+// @lcpr case=end
+
+// @lcpr case=start
+// [1]\n1\n
+// @lcpr case=end
+
+// @lcpr case=start
+// [1,2,1,2,1,2,3,1,3,2]\n2\n
+// @lcpr case=end
+
+ */