node.right = buildTree(nums,middle +1,end); node.sum = node.left.sum + node.right.sum; }returnnode; } }voidupdate(inti,intval){ update(i,root,val); }publicvoidupdate(inti,segmentNode root,intval){if(i == root.start && i == root.end) root.sum = val;else{intmiddle=root.start...
void update(int node, int start, int end, int index, int newValue) { if (start == end) { tree[node] = newValue; return; } int mid = (start + end) / 2; int lnode = node * 2; int rnode = node * 2 + 1; if (index <= mid) update(lnode, start, mid, index, new...
segmentTree.update(L, R, height, 1, N, 1); } return res; } 这里的查询和更新操作将求和变为了取最大值,整体的代码框架没有一点变化。 查询操作: public int query(int L, int R, int l, int r, int rt) { if (L <= l && r <= R) { return max[rt]; } int mid = (l + r) ...
假设闭包表结构为:<em>(简化,后续用 <code>节点指代的名称</code> 代替 <code>节点ID</code>)</em></p><pre><code class="sql">CREATE TABLE 闭包表 ( 祖先节点ID INT, 后代节点ID INT, 这俩节点距离 INT, PRIMARY KEY (祖先节点ID, 后代节点ID) );</code></pre><p>现有一张约 66W 行数据...
{ tree[u].right = ++idx; tree[tree[u].right] = new Node(); } } void pushup(int u) { tree[u].val = tree[tree[u].left].val + tree[tree[u].right].val; } public int ping(int t) { update(1, 1, N, t, 1); return query(1, 1, N, Math.max(0, t - 3000), t)...
Trie, Segment Tree,树状数组---Other Kinds of Tree 字典树(前缀树): Leetcode208 Implement Trie: class Trie { TrieNode root; //attribute, which can show a trie's identity /** Initialize your data structure here. */ public Trie() { //constuctor, which used to create a new trie instan...
updateRange(node->right, a, b, val); }SegTreeNode* root; vector<vector<int>>arr; } vector<pair<int,int>>height; // {idx, h} public:vector<vector<int>> getSkyline(vector<vector<int>>& buildings) { set<int>Set; for (auto& building: buildings) ...
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'''returnself.query(0,0,self.n-1,ql,qr)defbuild(self,tree_index:int,l:int,r:int):''' 递归创建线段树 tree_index : 线段树节点在数组中位置 l, r : 该节点表示的区间的左,右边界 '''ifl==r:self.tree[tree_index]=self.data[l]returnmid=(l+r)//2# 区间中点,对应左孩子区间结束,右孩...
这个算法的缺点是什么呢?Query询问最大值复杂度O(N), 修改复杂度为O(1),在有Q个query的情况下这样总的复杂度为O(QN), 那么对于查询来说这样的复杂度是不可以接受的。 Segment Tree 线段树 一颗线段树的构造就是根据区间的性质的来构造的, 如下是一棵区间[0, 3]的线段树,每个[start, end]都是一个二叉树...