LeetCode 327. Count of Range Sum-优快云博客

本文介绍了一个高效的算法，用于计算整数数组中所有范围和落在指定区间内的数量。该算法使用了分治法和二分查找技术，避免了朴素算法的O(n²)复杂度，通过预处理前缀和及排序过程中的巧妙计数实现了更优的时间复杂度。

Problem Statement

(Source) 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.

Tags: Divide and Conquer, Binary Search Tree.

Solution

class Solution(object):
    def countRangeSum(self, nums, lower, upper):
        """
        :type nums: List[int]
        :type lower: int
        :type upper: int
        :rtype: int
        """
        pre_sum = [0]
        for num in nums:
            pre_sum.append(pre_sum[-1] + num)

        def help(first, lo, hi):
            mid = (lo + hi) >> 1
            if mid == lo:
                return 0
            count = help(first, lo, mid) + help(first, mid, hi)
            i = j = mid
            for left in first[lo : mid]:
                while i < hi and first[i] - left < lower: i += 1
                while j < hi and first[j] - left <= upper: j += 1
                count += j - i
            first[lo : hi]  = sorted(first[lo : hi])
            return count

        return help(pre_sum, 0, len(pre_sum))