keep going

.gitignore

+~$etcode总结.docx

amazon.cpp

@@ -1312,17 +1312,107 @@ public:
 
 
 
+Kth Largest Element in an Array
+// Selection sort 典型的运用，
+class Solution {
+public:
+    int findKthLargest(vector<int>& nums, int k) {
+        for(int outer=0; outer<nums.size(); outer++){
+            int max_pos = outer;
+            for(int inner=outer; inner<nums.size(); inner++){
+                if(nums[max_pos]<nums[inner]) max_pos = inner;
+            }
+            if(outer==k-1) return nums[max_pos];
+            swap(nums[outer], nums[max_pos]);
+        }
+        return -1;
+    }
+};
 
 
 
 
+Top K Frequent Elements
+//第一次用priority_queue。。。。当pq只考虑最大的top时，complexity为 O(1)，所以合适
+//考虑过用map<int, int>但map不支持多个相同frequency的出现呀！所以不能用！（一个key出现一次）
+//老规矩，要计数的就用 hash table！好用！
+class Solution {
+public:
+    vector<int> topKFrequent(vector<int>& nums, int k) {
+        // build the freq table
+        unordered_map<int, int> hash;
+        for(auto &num: nums) hash[num]++;
+        
+        //build the frequency table
+        priority_queue<pair<int, int>> pq;
+        for(auto it=hash.begin(); it != hash.end(); it++){
+           pq.push(make_pair(it->second, it->first));
+        }
+        
+        vector<int> res;
+        while(k>0){
+            k--;
+            res.push_back(pq.top().second);
+            pq.pop();
+        }
+        return res;
+    }
+};
 
 
 
+K Closest Points to Origin
+// 思路和top k frequent element 有点像
+// 也是用pq解，问题在这个compator有点难写。。。。。。。。。
+//solution2 用默认comparator，维护pq的size，也不失为一种方法
+class Solution {
+    struct compare {
+        bool operator()(pair<double, int> &a, pair<double, int> &b) {
+            return a.first > b.first;
+        }
+    };
+    
+public:
+    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
+        priority_queue<pair<double, int>, vector<pair<double, int>>, compare > pq;
+        for(int i=0; i<points.size(); i++){
+            double dist = (0-points[i][0]) * (0-points[i][0]) + (0-points[i][1]) * (0-points[i][1]);
+            pq.push(make_pair(dist, i));
+        }
+        
+        vector<vector<int>> res;
+        
+        while(K>0){
+            K--;
+            int index = pq.top().second;
+            pq.pop();
+            res.push_back(points[index]);
+        }
+        
+        return res;
+    }
+};
 
-
-
-
+// solution2
+class Solution {
+public:
+    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
+        priority_queue<pair<double, int>> pq;
+        for(int i=0; i<points.size(); i++){
+            double dist = (0-points[i][0]) * (0-points[i][0]) + (0-points[i][1]) * (0-points[i][1]);
+            pq.push(make_pair(dist, i));
+            if(pq.size()>K) pq.pop();
+        }
+        vector<vector<int>> res;
+        while(K>0){
+            K--;
+            int index = pq.top().second;
+            pq.pop();
+            res.push_back(points[index]);
+        }
+        return res;
+    }
+};