ii. getSuccessor(N), which returns the next larger node to N. 2. Try to assume that each node has a parent pointer, it makes the problem much easier. 3. Without parent pointer we just need to keep track of the path from the root to the current node using a stack. 4. You would ...
vector<vector<int>> res; priority_queue<pair<int,int>> pq;for(inti =0; i < points.size(); ++i) {intt = points[i][0] * points[i][0] + points[i][1] * points[i][1]; pq.push({t, i});if(pq.size() > K) pq.pop(); }while(!pq.empty()) {autot = pq.top(); pq...
As with any greedy algorithm, there's the subtle problem with the possibility of getting stuck in a local optimum. So what in particular exempts our "two-pointer from both ends" solution from this? It can always be a headache to mathematically give a proof, but, well, we have no choice...
O(nlogn) time http://www.lintcode.com/en/problem/subarray-sum-closest/ 题目的意思是在一个数组中找一段连续的区间,使得这段区间的和的绝对值最小。做法就是利用前缀和,先用一个数组acc[i]来保存从nums[0]到nums[i]的和,同时还要记录下标,所以这里我用pair<int, int>来保存。那么,我们想要得到nums[...
* and the index of the last number*/vector<int> subarraySumClosest(vector<int>nums) {if(nums.empty()) {return{}; }constintn =nums.size(); vector<pair<int,int>> psum(n +1);//Get and Sort the partial sum. a little greedyfor(inti =0; i < n; ++i) { ...
Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines the closest common ancestor of u and v in the tree. The closest common ancestor of two nodes u and v is the node w that is an ancestor of both u and ...