leetcode 307. Range Sum Query

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) -> 9update(1, 2)sumRange(0, 2) -> 8

Note:

1. The array is only modifiable by the update function.
2. 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