910. 最小差值 II Smallest Range II 力扣每日一题 LeetCode 题解 [贪心算法 隔板法] 05:36 3191. 使二进制数组全部等于 1 的最少操作次数 力扣每日一题 LeetCode 题解 [贪心算法 遍历] 05:22 3193. 统计逆序对的数目 Count the Number of Inversions 力扣每日一题 LeetCode 题解 [动态规划 递推 ...
**/intLeetCode::countRangeSum(vector<int>& nums,intlower,intupper){intn =nums.size(); vector<vector<int>> sum(n,vector<int>(n,0));intcount =0;for(size_t i =0; i < n; i++){ sum[i][i]=nums[i];if(sum[i][i] >= lower && sum[i][i] <= upper)++count; }for(size...
You are given an integer arraynumsand you have to return a newcountsarray. Thecountsarray has the property wherecounts[i]is the number of smaller elements to the right ofnums[i]. Example: Input: [5,2,6,1] Output:[2,1,1,0] Explanation:To the right of 5 there are 2 smaller elemen...
Example: Input: nums = [-2,5,-1], lower = -2, upper = 2, Output: 3 Explanation: The three ranges are : [0,0], [2,2], [0,2] and their respective sums are: -2, -1, 2. 解题思路: 1、累加nums元素置为sums列表中; 2、对于nums中的每个元素,需要找到其前面元素之差在范围内,即...
You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i]. Example: Input: [5,2,6,1] Output: [2,1,1,0] ...
The three ranges are : [0, 0], [2, 2], [0, 2] and their respective sums are: -2, -1, 2. 我没有想到更好的方法,直接循环做的。 建议和leetcode 307. Range Sum Query - Mutable 树状数组的一个应用 一起学习 代码如下: /*
You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i]. Example: Input: [5,2,6,1] Output: [2,1,1,0] ...
题目来自leetcode3072。 大体步骤如下: 1.创建一个新的函数greaterCount(arr, val),用于计算数组arr中大于val的元素数量。 2.定义一个空数组arr1和arr2,并创建两个BinaryIndexedTree数据结构tree1和tree2。 3.对于数组nums中的每个元素: 3.1. 将当前元素按照索引排序,并通过Binary Indexed Tree记录每个元素在排序...
select count(*) from order where status=0;但如果在一个接口中，同步执行这两条sql效率会非常低。...
The three ranges are : [0, 0], [2, 2], [0, 2] and their respective sums are: -2, -1, 2. 思路: range sum一般会转化为前缀和的问题prefix sum 解法(分治法): intcountRangeSum(vector<int>&nums,int lower,int upper){//get the prefix sumvector<long long>prefix(nums.size()+1,0)...