LintCode 249. 統計前面比自己小的數的個數
阿新 • • 發佈:2018-04-07
自己 treenode 感覺 num ini 求和 內置 tor -o
給定一個整數數組(下標由 0 到 n-1, n 表示數組的規模,取值範圍由 0 到10000)。對於數組中的每個 ai
元素,請計算 ai
前的數中比它小的元素的數量。
註意事項
We suggest you finish problem Segment Tree Build, Segment Tree Query II and Count of Smaller Number first.
樣例
對於數組[1,2,7,8,5]
,返回 [0,1,2,3,2]
解題思路:
題目提示我們使用線段樹,在這裏我寫了兩種線段樹的解法,一種TLE,一種正常通過;個人感覺兩種寫法需要因地制宜使用:
思路1:每個線段樹節點存儲的是原始vector的index前後值以及該區間內的相應最大值,在查詢時,通過區域以及最大值約束找到所有小於特定值的區間最後求和。
class SegmentTreeNode{ //線段樹節點,其中max是當前區域內(left-right)最大值 public: int start,end,max; SegmentTreeNode2 * left; SegmentTreeNode2 * right; SegmentTreeNode2(int x,int y,int max){ this->start = x; this->end = y; this->max = max; this->left = this->right = nullptr; } }; class Solution { public: /** * @param A: an integer array * @return: A list of integers includes the index of the first number and the index of the last number */ vector<int> countOfSmallerNumberII(vector<int> &A) { // write your code here auto tree = new SegmentTreeNode(0,A.size()-1,INT_MIN); buildTree(A,tree); vector<int> ret; for(int i = 0;i<A.size();++i){ ret.push_back(query(tree,0,i,A[i])); } return ret; } int buildTree(vector<int> &A,SegmentTreeNode * root){ //建立線段樹,每個節點保存該區域內最大值 int start = root->start; int end = root->end; if(start > end) return 0; if(start == end) { root->max = A[start]; return A[start]; }else{ root->left = new SegmentTreeNode(start,(start+end)/2,INT_MIN); root->right = new SegmentTreeNode((start+end)/2+1,end,INT_MIN); int L_max = buildTree(A,root->left); int R_max = buildTree(A,root->right); root->max = L_max>R_max?L_max:R_max; return root->max; }; } int query(SegmentTreeNode * root,int start,int end,int q){ //查詢特定區域比q小的個數 if(root == nullptr) return 0; if(root->start > end || root->end < start) return 0; if(root->start >= start && root->end <= end && root->max<q) return root->end - root->start + 1; return query(root->left,start,end,q)+query(root->right,start,end,q); } };
這種解法TLE,時間復雜度在vector的size很大時很大,某種程度上來講效率不及直接暴力法,但當所求數據較為集中時應該能提高一點速度。
思路2:對數據的區間建立線段樹,在知道所有數據上界的情況下效率不錯,能夠正常通過
class SegmentTreeNode{//count表示當前區間所有的數個數 public: int start,end,count; SegmentTreeNode * left; SegmentTreeNode * right; SegmentTreeNode(int x,int y,int count){ this->start = x; this->end = y; this->count = count; this->left = this->right = nullptr; } }; class Solution { public: /** * @param A: an integer array * @return: A list of integers includes the index of the first number and the index of the last number */ vector<int> countOfSmallerNumberII(vector<int> &A) { // write your code here vector<int> res; SegmentTreeNode * root = buildTree(0,10001); for(int i=0; i<A.size(); ++i){ res.push_back(query(root,A[i])); insert(root,A[i]); } return res; } SegmentTreeNode* buildTree(int start,int end){ //這種方法需要明確數據上界,然後直接根據數據大小建立線段樹,每個節點保存落在當前區間數的個數 if(start > end) return nullptr; auto res = new SegmentTreeNode(start,end,0); if(start == end) return res; int mid = (start+end)/2; res->left = buildTree(start,mid); res->right = buildTree(mid+1,end); return res; } int query(SegmentTreeNode * root,int q){ //query函數用來查詢當前區域內小於q的數的個數 if(root == nullptr) return 0; if(q < root->start) return 0; if(q > root->end) return root->count; return query(root->left,q)+query(root->right,q); } void insert(SegmentTreeNode * root,int val){//將輸入數據逐個插入,從上到下逐個更新count if(root == nullptr) return; if(root->start > val || root->end < val) return; if(root->start <= val && root->end >= val) ++root->count; insert(root->left,val); insert(root->right,val); } }
ps:這道題如果使用lintcode內置的SegmentTreeNode 數據結構中的count好像會出問題,最好定義自己的class
LintCode 249. 統計前面比自己小的數的個數