leetcode 307. Range Sum Query - Mutable

307. Range Sum Query - Mutable

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.

1、下面是自己开始写的原始方式。

2、后面有学习的线段树的方式。

class NumArray {
public:
NumArray(vector<int> nums)
{
if (!nums.empty())
{
num = nums;
sum.push_back(num[0]);
for (int i = 1; i < num.size(); i++)
sum.push_back(sum[i - 1] + num[i]);
}
}

void update(int i, int val)
{

for (int k = i; k < num.size(); k++)
{
sum[k] += val - num[i];
}
num[i] = val;
}

int sumRange(int i, int j)
{
if (i == 0) return sum[j];
return sum[j] - sum[i - 1];
}
private:
vector<int> sum; //前n项和
vector<int> num;
};

/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* obj.update(i,val);
* int param_2 = obj.sumRange(i,j);
*/

2、线段树

class NumArray {
public:
NumArray(vector<int> nums)
{
if (!nums.empty())
{
Num = nums;
for (int i = 0; i < Num.size() * 4; i++)
TreeSum.push_back(0);
Build(1, 0, Num.size() - 1);
}
}

/*i表示当前递归编号,l,r分别表示当前点的左右区间*/
/* Tree数组是存储 线段树的数组 */
void Build(int i, int l, int r)
{
if (l == r)
{
TreeSum[i] = Num[l];
return;
}
int Mid = (l + r) / 2;
Build(i * 2, l, Mid);
Build(i * 2 + 1, Mid + 1, r);
PushUp(i);  //下面的子节点，需要把 和 传上来
}

void update(int i, int val)
{
Update_Single(1, 0, Num.size() - 1, i, val);
}

/*i为当前编号,L,R为左右区间,A为修改点的编号,B为修改的值*/
void Update_Single(int i, int L, int R, int A, int B)
{
if (L == R)
{
/*如果找到了,修改值*/
TreeSum[i] = B;
return;
}
int Mid = (L + R) / 2;
if  (A <= Mid)
Update_Single(i * 2, L, Mid, A, B); /*递归左子树*/
else
Update_Single(i * 2 + 1, Mid + 1, R, A, B); /*递归右子树*/
PushUp(i); //下面的子节点，需要把 修改后的和 传上来
}

int sumRange(int i, int j)
{
return Quary_Total(1, 0, Num.size() - 1, i, j);
}

/*i 为当前编号, L, R为当前i节点的区间。 l、r 是使用者查询的区间*/
int Quary_Total(int i, int L, int R, int l, int r)
{
if (l == L && r == R) return TreeSum[i]; /*如果刚好是在当前区间*/
int Mid = (L + R) / 2;
if(l <= Mid && r > Mid)  //查询区间 在当前区间 的 左右两小部分 都有的情况
return Quary_Total(i * 2, L, Mid, l, Mid) + Quary_Total(i * 2 + 1, Mid + 1, R, Mid + 1, r);
else if (r <= Mid) //只在左边部分
return Quary_Total(i * 2, L, Mid, l, r); /*递归左子树*/
else if (l > Mid)  //只在右边部分
return Quary_Total(i * 2 + 1, Mid + 1, R, l, r); /*递归右子树*/
}

/*区间和处理*/
void PushUp(int Now)
{
TreeSum[Now] = TreeSum[Now * 2] + TreeSum[Now * 2 + 1];
}

private:
vector<int> TreeSum;  //注意 TreeSum编号是从1开始的
vector<int> Num;    //只在最开始build的时候有用
};

/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* obj.update(i,val);
* int param_2 = obj.sumRange(i,j);
*/

https://www.cnblogs.com/xiaoyao24256/p/6590885.html 线段树参考的是这篇博客