对于一个数组,我们可以对其建立一棵 线段树, 每个结点存储一个额外的值 count 来代表这个结点所指代的数组区间内的元素个数. (数组中并不一定每个位置上都有元素)
实现一个 query 的方法,该方法接受三个参数 root, start 和 end, 分别代表线段树的根节点和需要查询的区间,找到数组中在区间[start, end]内的元素个数。
注意事项:
It is much easier to understand this problem if you finished Segment Tree Buildand Segment Tree Query first.
样例:
对于数组 [0, 空,2, 3], 对应的线段树为:
[0, 3, count=3]
/ \
[0,1,count=1] [2,3,count=2]
/ \ / \
[0,0,count=1] [1,1,count=0] [2,2,count=1], [3,3,count=1]
query(1, 1), return 0
query(1, 2), return 1
query(2, 3), return 2
query(0, 2), return 2
#ifndef C247_H
#define C247_H
#include<iostream>
using namespace std;
class SegmentTreeNode{
public:
int start, end, count;
SegmentTreeNode *left, *right;
SegmentTreeNode(int start, int end, int count){
this->start = start;
this->end = end;
this->count = count;
this->left = this->right = NULL;
}
};
class Solution {
public:
/*
* @param root: The root of segment tree.
* @param start: start value.
* @param end: end value.
* @return: The count number in the interval [start, end]
*/
int query(SegmentTreeNode *root, int start, int end) {
// write your code here
if (start > end||root==NULL)
return 0;
if (start < root->start)
return query(root, root->start, end);
if (end>root->end)
return query(root, start, root->end);
SegmentTreeNode *node = root;
if (start >= node->start&&end <= node->end&&node->count == 0)
return 0;
if (start == node->start&&end == node->end)
return node->count;
if (end <= node->left->end)
{
return query(node->left, start, end);
}
else if (start >= node->right->start)
{
return query(node->right, start, end);
}
else
{
int l = query(node->left, start, node->left->end);
int r = query(node->right, node->right->start, end);
return l + r;
}
}
};
#endif