Notes

tags: Priority Queue,LeetCode101 Using a min heap to keep k elements, top is the Kth largest element. class KthLargest { private: // min heap priority_queue<int, vector<int>, std::greater<int>> pq; int K; public: KthLargest(int k, vector<int>& nums) { K = k; for (auto n: nums) { add(n); } } int add(int val) { pq.push(val); while (pq.size() > K) { pq.pop(); } return pq.top(); } }; /** * Your KthLargest object will be instantiated and called as such: * KthLargest* obj = new KthLargest(k, nums); * int param_1 = obj->add(val); */

tags: Heap (data structure),LeetCode101 Find top k values, classical heap problems. Here we need find top 2 values, and pop them from heap, then we will meet two cases: If they are not same, put back the differ between them, continue. Otherwise, continue directly.

tags: LeetCode101,Sorting,Two Pointers Key ideas: Move both sides to inwards. If the sum value less than target, move left pointer. Otherwise move right poinger. class Solution { public: vector<int> twoSum(vector<int>& numbers, int target) { int S, l = 0, r = numbers.size() - 1; vector<int> res(2, 0); while (l < r) { S = numbers[l] + numbers[r]; if (S == target) { break; } if (S < target) { l++; } else { r--; } } res[0] = l + 1; res[1] = r + 1; return res; } };

tags: LeetCode101,Sorting,Two Pointers,Three Pointers Key ideas see LeetCode101: 15. 3Sum class Solution { public: int threeSumClosest(vector<int>& nums, int target) { sort(nums.begin(), nums.end()); int closest = INT_MAX, l, r, sum, T, res; for (int i = 0; i < nums.size(); i++) { l = i + 1; r = nums.size() - 1; T = target - nums[i]; while (l < r) { sum = nums[l] + nums[r]; if (abs(sum - T) < closest) { res = sum + nums[i]; closest = abs(sum - T); } if (sum == T) { return target; } if (sum < T) { l++; } else { r--; } } } return res; } };

tags: LeetCode101,Sorting,Two Pointers,Three Pointers Key ideas: Sort the nums first. Then, we travel the nums, pick current element as nums[i], and apply LeetCode101: 167. Two Sum II - Input Array Is Sorted to the remains. We skip the same numbers to avoid duplicate. class Solution { public: vector<vector<int>> threeSum(vector<int>& nums) { sort(nums.begin(), nums.end()); int l, r, sum, T; vector<vector<int>> res; for (int i = 0; i < nums.size(); i++) { // Skip same numbers to avoid duplicate if(i > 0 && nums[i] == nums[i-1]) { continue; } l = i + 1; r = nums.size() - 1; T = 0 - nums[i]; while (l < r) { sum = nums[l] + nums[r]; if (sum == T) { res.push_back({nums[i], nums[l], nums[r]}); // Skip same numbers in each side to avoid duplicate while (l < r && nums[l] == nums[l + 1]) l++; while (l < r && nums[r] == nums[r - 1]) r--; l++; r--; } else if (sum < T) { l++; } else { r--; } } } return res; } };