【Leetcode每日一題】327. 區間和的個數（線段樹/樹狀陣列）

Leetcode 每日一題
題目連結： 327. 區間和的個數
難度： 困難
解題思路： 今天有急事，來不及做了。大概看了一下，區間問題，要求時間複雜度小於O(n*n)。很容易想到用線段樹或者樹狀陣列，搬運Leetcode 官方題解。
題解：

struct SegNode {
    int lo, hi, add;
    SegNode* lchild, *rchild;
    SegNode(int left, int right): lo(left), hi(right), add(0), lchild(nullptr), rchild(nullptr) {}
};

class Solution {
public:
    SegNode* build(int left, int right) {
        SegNode* node = new SegNode(left, right);
        if (left == right) {
            return node;
        }
        int mid = (left + right) / 2;
        node->lchild = build(left, mid);
        node->rchild = build(mid + 1, right);
        return node;
    }

    void insert(SegNode* root, int val) {
        root->add++;
        if (root->lo == root->hi) {
            return;
        }
        int mid = (root->lo + root->hi) / 2;
        if (val <= mid) {
            insert(root->lchild, val);
        }
        else {
            insert(root->rchild, val);
        }
    }

    int count(SegNode* root, int left, int right) const {
        if (left > root->hi || right < root->lo) {
            return 0;
        }
        if (left <= root->lo && root->hi <= right) {
            return root->add;
        }
        return count(root->lchild, left, right) + count(root->rchild, left, right);
    }

    int countRangeSum(vector<int>& nums, int lower, int upper) {
        long long sum = 0;
        vector<long long> preSum = {0};
        for (int v: nums) {
            sum += v;
            preSum.push_back(sum);
        }
        
        set<long long> allNumbers;
        for (long long x: preSum) {
            allNumbers.insert(x);
            allNumbers.insert(x - lower);
            allNumbers.insert(x - upper);
        }
        // 利用雜湊表進行離散化
        unordered_map<long long, int> values;
        int idx = 0;
        for (long long x: allNumbers) {
            values[x] = idx;
            idx++;
        }

        SegNode* root = build(0, values.size() - 1);
        int ret = 0;
        for (long long x: preSum) {
            int left = values[x - upper], right = values[x - lower];
            ret += count(root, left, right);
            insert(root, values[x]);
        }
        return ret;
    }
};

作者：LeetCode-Solution
連結：https://leetcode-cn.com/problems/count-of-range-sum/solution/qu-jian-he-de-ge-shu-by-leetcode-solution/
來源：力扣（LeetCode）