[LeetCode] Count of Range Sum 区间和计数

Posted 2020-06-09

 

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.

Example:
Given nums = [-2, 5, -1], lower = -2, upper = 2,
Return 3.
The three ranges are : [0, 0], [2, 2], [0, 2] and their respective sums are: -2, -1, 2.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

 

 

解法一：

class Solution {
public:
    int countRangeSum(vector<int>& nums, int lower, int upper) {
        int res = 0;
        long long sum = 0;
        multiset<long long> sums;
        sums.insert(0);
        for (int i = 0; i < nums.size(); ++i) {
            sum += nums[i];
            res += distance(sums.lower_bound(sum - upper), sums.upper_bound(sum - lower));
            sums.insert(sum);
        }
        return res;
    }        
};

 

 

 

解法二：

class Solution {
public:
    int countRangeSum(vector<int>& nums, int lower, int upper) {
        vector<long> sums(nums.size() + 1, 0);
        for (int i = 0; i < nums.size(); ++i) {
            sums[i + 1] = sums[i] + nums[i];
        }
        return countAndMergeSort(sums, 0, sums.size(), lower, upper);
    }
    int countAndMergeSort(vector<long> &sums, int start, int end, int lower, int upper) {
        if (end - start <= 1) return 0;
        int mid = start + (end - start) / 2;
        int cnt = countAndMergeSort(sums, start, mid, lower, upper) + countAndMergeSort(sums, mid, end, lower, upper);
        int j = mid, k = mid, t = mid;
        vector<int> cache(end - start, 0);
        for (int i = start, r = 0; i < mid; ++i, ++r) {
            while (k < end && sums[k] - sums[i] < lower) ++k;
            while (j < end && sums[j] - sums[i] <= upper) ++j;
            while (t < end && sums[t] < sums[i]) cache[r++] = sums[t++];
            cache[r] = sums[i];
            cnt += j - k;
        }
        copy(cache.begin(), cache.begin() + t - start, sums.begin() + start);
        return cnt;
    }
};

 

参考资料：

https://leetcode.com/discuss/79083/share-my-solution

https://leetcode.com/discuss/79632/multiset-solution-100ms-binary-search-tree-180ms-mergesort

https://leetcode.com/discuss/79907/summary-divide-conquer-based-binary-indexed-based-solutions

 

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