LeetCode 307. Range Sum Query - Mutable

最新推荐文章于 2020-05-20 18:29:12 发布

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最新推荐文章于 2020-05-20 18:29:12 发布

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分类专栏： LeetCode 题解 Multiple Passes

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本文介绍了一种使用二叉索引树(Binary Index Tree, BIT)进行高效区间求和及更新的方法。通过实例展示了二叉索引树的组织结构，并详细解释了如何通过更新和获取操作来维护树的数据。

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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.

#include <vector>
#include <iostream>
using namespace std;
// this is actually to use binary index tree.
int getSum(vector<int>& BITree, int index) {
  int sum = 0;
  index = index + 1;
  while(index > 0) {
    sum += BITree[index];
    index -= index & (-index); // update to its parent node in getSum view.
  }
  return sum;
}

void updateBITree(vector<int>& BITree, int index, int val) {
  index = index + 1;
  while(index < BITree.size()) {
    BITree[index] += val;
    index += index & (-index); // update to its parent node in update view
  }
}

vector<int> createBITree(vector<int>& nums) {
  int n = nums.size();
  vector<int> BITree(n + 1, 0);
  for(int i = 0; i < n; ++i) {
    updateBITree(BITree, i, nums[i]);
  }
  return BITree;
}

int main(void) {
  vector<int> nums{1, 3, 5};
  vector<int> BITree = createBITree(nums);
  updateBITree(BITree, 1, 2);
  cout << getSum(BITree, 2) << endl;
}

Binary Index Tree: See explain here: Binary Index Tree However, how the index is organized is not clear.

Let's use an example to help explain how to organize a binary index tree.

Suppose we have a 10 size array [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Our target tree will be like this: