###问题描述 给定一个整数数组,返回range sum 落在给定区间[lower, upper] (包含lower和upper)的个数.range sum S(i, j) 表示数组中第i 个元素到j 个元素之和. Note: A naive algorithm of O(n2) is trivial. You MUST do better than that. Example: Input: nums = [-2,5,-1], lower = -2, upper = 2, Output: 3…

问题描述: 给定一个整数数组nums,返回其所有落在[low, upper]范围内(包含边界)的区间和的数目. 区间和sums(i, j)的定义为所有下标为i到j之间(i ≤ j)的元素的和,包含边界. 思路: 归并排序 问题可描述为,对每个sums[i],满足lower<=sums[i] - sums[j]<=upper的j的个数,并进行求和. 一个很巧妙的方法是,对sums进行排序得到ocums,然后求ocums[m] - ocums[n] = sums[i] - sums[j]落在上下界…

又是一道有意思的题目,Count of Range Sum.(PS:leetcode 我已经做了 190 道,欢迎围观全部题解 https://github.com/hanzichi/leetcode) 题意非常简单,给一个数组,如果该数组的一个子数组,元素之和大于等于给定的一个参数值(lower),小于等于一个给定的参数值(upper),那么这为一组解,求总共有几组解. 一个非常容易想到的解法是两层 for 循环遍历子数组首尾,加起来判断,时间复杂度 O(n^2). /** * @param…

/* * 327. Count of Range Sum * 2016-7-8 by Mingyang */ public int countRangeSum(int[] nums, int lower, int upper) { int n = nums.length; long[] sums = new long[n + 1]; for (int i = 0; i < n; ++i) sums[i + 1] = sums[i] + nums[i]; return countWhileMerg…

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 numsbetween indices i and j (i ≤ j), inclusive. Note:A naive algorithm of O(n2) is trivial. Y…

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.…

原题链接在这里:https://leetcode.com/problems/count-of-range-sum/ 题目: 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…

https://leetcode.com/problems/count-of-range-sum/ 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), incl…

题目: 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 tr…

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.…