这一题,我们使用了分治法. 首先看时间复杂度为o(n^2),比较naïve的方法: 使用一个数组sum[],长度为原数组长度+1,每一个元素sum[i]为原数组中index0~i-1之和,显然当sum[j] – sum[i]就是i~j-1之和,于是我们只需要两个for来遍历所有[i, j],并且比较是否在lower~upper之间即可. 但是题目要求时间复杂度由于o(n^2),考虑使用分治 1) 分:将sum[]分为左右两部分,分别计算满足条件的[i,j] 2) 合:除了i,j都在左或者都在右,…

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…

Give you an integer array (index from 0 to n-1, where n is the size of this array, value from 0 to 10000) and an query list. For each query, give you an integer, return the number of element in the array that are smaller than the given integer. Have…

Almost identical to LintCode "Count of Smaller Number before Self". Corner case needs to be taken care of. class Solution { ////////////////// // Fenwick Tree // vector<long long> ft; void update(int i, long long x) { ) { ft[] ++; return;…

[题目描述] Give you an integer array (index from 0 to n-1, where n is the size of this array, data value from 0 to 10000) . For each element Ai in the array, count the number of element before this element Ai is smaller than it and return count number ar…

/* * 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 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…