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.…

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. 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.…

/* * 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…

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…

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.…

无意看到的LeetCode新题,不算太简单,大意是给一个数组,询问多少区间和在某个[L,R]之内.首先做出前缀和,将问题转为数组中多少A[j]-A[i] (j>i)在范围内. 有一种基于归并排序的做法,在每次归并完左右两个子区间后,当前区间两部分分别都已经排序完毕,基于有序这一点,扫描后半段区间,对于每个A[i] (i>=mid),目标区间即为[ A[i]-R, A[i]-L ], 对于有序数组来说,求出元素落在某一区间的个数直接就是upper_bound-lower_bound,事实上,这里…

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…