327. Count of Range Sum Hard-优快云博客

本文链接：https://blog.youkuaiyun.com/wakakon/article/details/72021100

本文介绍了一种有效的区间和统计算法，该算法通过归并排序优化复杂度，实现快速计算数组中满足条件的区间和数量。文章详细展示了如何利用数列和的概念将问题转化为数列和的差，并通过分治思想进一步提升效率。

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Given an integer array nums, return the number of range sums that lie in [lower, upper] inclusive.
Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j (i ≤ j), inclusive.

Note:

A naive algorithm of O(n2) is trivial. You MUST do better than that.

思路：用一个数组存储一个数列和，这些和为nums中从nums[0]开始以各项数为终点的和，这样要求的子数列和就转化为这些数列和的差的形式，枚举这些差就可以得到满足区间类的和的个数。考虑一个特殊情况，当数列和是单调递增的情况下，复杂度有所下降，因此我们使用分治中的归并排序来降低复杂度。

int range(vector<long long int>&sum, int start, int end, int upper, int lower) {
    if (start >= end) {
        return 0;
    }
    int mid = (end - start) / 2;
    mid += start;
    int count = range(sum, start, mid, upper, lower) + range(sum, mid + 1, end, upper, lower);
    vector<long long int>record(end - start + 1, 0);
    int q = mid + 1, k = q, j = q, m = 0;
    for (int i = start; i <= mid; i++, m++) {
        while(q <= end && sum[q] - sum[i] <= upper) {
            q++;
        }
        while (k <= end && sum[k] - sum[i] < lower) {
            k++;
        }
        count += q - k;
        while (j <= end && sum[j] <= sum[i]) {
            record[m] = sum[j];
            m++;
            j++;
        }
        record[m] = sum[i];
    }  
    for (int i = 0; i < m; i++) {
        sum[start + i] = record[i];
    }
    return count;
}

class Solution {
public:
    int countRangeSum(vector<int>& nums, int lower, int upper) {
        if (nums.size() == 0) {
            return 0;
        }
        vector<long long int>sum(nums.size() + 1, 0);
        for (int i = 0; i < nums.size(); i++) {
            sum[i + 1] =  sum[i] + nums[i];
        }
        return range(sum, 0, nums.size(), upper, lower);
    }
};