线段树
线段树是一棵树，而且是二叉搜索树。它将一个区间划分成一些单元区间，每个单元区间对应线段树中的一个叶结点。
主要应用：适用于和区间统计有关的问题，例如大数据的动态修改 及多次查询就比较适合使用这种树，效果比较好。
性质：
1. 树中的每一个结点表示了一个区间[a,b]。a,b通常是整数。
2. 对于每一个非叶结点所表示的结点[a,b]，其左儿子表示的区间为
[a,(a+b)/2]，右儿子表示的区间为[(a+b)/2,b]。
3. 同一层的节点所代表的区间，相互不会重叠。
4. 叶子节点的区间是单位长度，不可再分。
构造过程：
1. 构造节点node,对应的区间为[start,end]
2. 构造node的左孩子，对应的区间为[start,start + (end - start) / 2]
3. 构造node右孩子，对应的区间为[start + (end - start) / 2 + 1,end]
线段树的构造过程就是二叉树递归的创建的过程，代码如下：
SegmentTreeNode* SegmentTree::GenerateTree(vector<int> &nums, int start, int end) {
if(start > end) return nullptr;
SegmentTreeNode *node = new SegmentTreeNode(start,end);
if(start == end) {
node->SetSum(nums[start]);
return node;
}
int mid = start + (end - start) / 2;
SegmentTreeNode *left = GenerateTree(nums,start,mid);
SegmentTreeNode *right = GenerateTree(nums,mid+1,end);
node->SetLeft(left);
node->SetRight(right);
node->SetSum(left->GetSum() + right->GetSum());
return node;
}
其中SegmentTreeNode 以及 SegmentTree为自己定义的类，最后附上整个的代码。
#include <vector>
using namespace std;
class SegmentTreeNode {
public:
SegmentTreeNode(int a, int b);
int GetStart();
int GetEnd();
void SetSum(int sum);
int GetSum();
void SetLeft(SegmentTreeNode* left);
SegmentTreeNode* GetLeft();
void SetRight(SegmentTreeNode* right);
SegmentTreeNode* GetRight();
private:
int start, end, sum;
SegmentTreeNode* left;
SegmentTreeNode* right;
};
class SegmentTree {
public:
void BuildTree(vector<int> &nums, int start, int end);
SegmentTreeNode *GetRoot();
int ModifyTree(int i, int val);
int QueryTree(int i, int j);
private:
SegmentTreeNode *GenerateTree(vector<int> &nums, int start, int end);
int _ModifyTree(int i, int val,SegmentTreeNode *node);
int _QueryTree(int i, int j,SegmentTreeNode *node);
SegmentTreeNode *root;
};
SegmentTreeNode::SegmentTreeNode(int a, int b):start(a),
end(b),
sum(0),
left(nullptr),
right(nullptr) {
};
int SegmentTreeNode::GetStart() {
return start;
}
int SegmentTreeNode::GetEnd() {
return end;
}
void SegmentTreeNode::SetSum(int sum) {
this->sum = sum;
}
int SegmentTreeNode::GetSum() {
return sum;
}
void SegmentTreeNode::SetLeft(SegmentTreeNode* left) {
this->left = left;
}
SegmentTreeNode* SegmentTreeNode::GetLeft() {
return left;
}
void SegmentTreeNode::SetRight(SegmentTreeNode* right) {
this->right = right;
}
SegmentTreeNode* SegmentTreeNode::GetRight() {
return right;
}
void SegmentTree::BuildTree(vector<int> &nums, int start, int end) {
root = GenerateTree(nums, start, end);
}
SegmentTreeNode* SegmentTree::GenerateTree(vector<int> &nums, int start, int end) {
if(start > end) return nullptr;
SegmentTreeNode *node = new SegmentTreeNode(start,end);
if(start == end) {
node->SetSum(nums[start]);
return node;
}
int mid = start + (end - start) / 2;
SegmentTreeNode *left = GenerateTree(nums,start,mid);
SegmentTreeNode *right = GenerateTree(nums,mid+1,end);
node->SetLeft(left);
node->SetRight(right);
node->SetSum(left->GetSum() + right->GetSum());
return node;
}
int SegmentTree::ModifyTree(int i, int val) {
return _ModifyTree(i, val, root);
}
int SegmentTree::QueryTree(int i, int j) {
return _QueryTree(i, j, root);
}
int SegmentTree::_ModifyTree(int i, int val,SegmentTreeNode *node) {
if(node == nullptr) return 0;
int diff;
if(node->GetStart() == i && node->GetEnd() == i) {
diff = val - node->GetSum();
node->SetSum(val);
return diff;
}
int mid = (node->GetStart() + node->GetEnd()) / 2;
if(i > mid) {
diff = _ModifyTree(i,val,node->GetRight());
} else {
diff = _ModifyTree(i,val,node->GetLeft());
}
node->SetSum(node->GetSum() + diff);
return diff;
}
int SegmentTree::_QueryTree(int i, int j,SegmentTreeNode *node) {
if(node == nullptr) return 0;
if(node->GetStart() == i && node->GetEnd() == j) return node->GetSum();
int mid = (node->GetStart() + node->GetEnd()) / 2;
if(i > mid) return _QueryTree(i,j,node->GetRight());
if(j <= mid) return _QueryTree(i,j,node->GetLeft());
return _QueryTree(i,mid,node->GetLeft()) + _QueryTree(mid+1,j,node->GetRight());
}
利用此代码在https://leetcode.com上AC过了两道题目。
Question 1: Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
Example: Given nums = [-2, 0, 3, -5, 2, -1]
sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You may assume that the array does not change. There are many calls to sumRange function.
Question 2: Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
The update(i, val) function modifies nums by updating the element at index i to val. Example: Given nums = [1, 3, 5]
sumRange(0, 2) -> 9 update(1, 2) sumRange(0, 2) -> 8 Note: The array is only modifiable by the update function. You may assume the number of calls to update and sumRange function is distributed evenly.
源码的下载地址：https://github.com/FyhSky/SegmentTree