Range Sum Query - Mutable-优快云博客

本文介绍了一种使用线段树实现数组元素实时更新及区间求和的方法。通过构建特殊的线段树结构，可以高效地完成区间求和操作，并支持对数组任意位置的值进行更新。具体实现中，利用了数组来存储线段树节点，简化了数据结构的维护。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem

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

Solution

自己创建一个segment tree的数据结构老是超时。。

参考了官方的答案，用数组来实现一个数，挺巧妙的。

class NumArray {
    int N;
    vector<int> tree;
    int find( int left, int right) {
        if(left > right) return 0;
        if(left == right) return tree[left];
        int sum = 0;
        if( left%2 == 1 ){
            sum += tree[left++];
        } 
        if(right%2 == 0 ) {
            sum += tree[right--];
        }
        return sum += find( left/2, right/2);
    }
public:
    NumArray(vector<int> &nums): N(nums.size()) {
        
        tree.resize(2*N,0);
        for( int i = N; i < 2*N; i++){
            tree[i] = nums[i-N];
        }
        for( int i = N - 1; i > 0; i--){
            tree[i] = tree[2*i] + tree[2*i+1];
        }
    }

    void update(int idx, int val) {
        idx += N;
        int diff = val - tree[idx];
        while(idx != 0) {
            tree[idx] += diff;
            idx /= 2;
        }
    }

    int sumRange(int i, int j) {
        return find( i + N, j + N);
    }
};


// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.update(1, 10);
// numArray.sumRange(1, 2);