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:[5,2,6,1] Explanation: To the…

315. Count of Smaller Numbers After Self class Solution { public: vector<int> countSmaller(vector<int>& nums) { int n = nums.size(); vector<int> v(n); for (int i = n - 1; i >= 0; --i) { int val = nums[i]; int L = i + 1, R = n - 1;…

说来惭愧,已经四个月没有切 leetcode 上的题目了. 虽然工作中很少(几乎)没有用到什么高级算法,数据结构,但是我一直坚信 "任何语言都会过时,只有数据结构和算法才能永恒".leetcode 上的题目,截止目前切了 137 道(all solutions),只写过 6 篇题解,所以我会写题解的一般都是自认为还蛮有意思或者蛮典型的题目,就比如这道题. 题目链接:Count of Smaller Numbers After Self 这道题很有意思,给出一个数组,返回一个新的数组,新…

说来惭愧,已经四个月没有切 leetcode 上的题目了. 虽然工作中很少(几乎)没有用到什么高级算法,数据结构,但是我一直坚信 "任何语言都会过时,只有数据结构和算法才能永恒".leetcode 上的题目,截止目前切了 137 道(all solutions),只写过 6 篇题解,所以我会写题解的一般都是自认为还蛮有意思或者蛮典型的题目,就比如这道题. 题目链接:Count of Smaller Numbers After Self 这道题很有意思,给出一个数组,返回一个新的数组,新…

You are given an integer array nums and you have to return a new counts array. The countsarray 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] Explanation: To the…

原题链接在这里:https://leetcode.com/problems/count-of-smaller-numbers-after-self/ 题目: 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 r…

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: Given nums = [5, 2, 6, 1] To the right of 5 there are…

You are given an integer array nums and you have to return a new counts array. Thecounts array has the property where counts[i] is the number of smaller elements to the right of nums[i]. Example: Given nums = [5, 2, 6, 1] To the right of 5 there are…

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 1: Input: nums = [5,2,6,1] Output: [2,1,1,0] Explanati…

Leetcode315 题意很简单,给定一个序列,求每一个数的右边有多少小于它的数. O(n^2)的算法是显而易见的. 用普通的线段树可以优化到O(nlogn) 我们可以直接套用主席树的模板. 主席树的功能是什么呢? 其实就是一句话. 原序列a的子序列a[l,r]在a排序后的序列b的子序列[L,R]中的个数. 显然本题只用到了主席树的一小部分功能. const int N = 100000 + 5; int a[N], b[N], rt[N * 20], ls[N * 20], rs[N * 2…

给定一个整型数组 nums,按要求返回一个新的 counts 数组.数组 counts 有该性质: counts[i] 的值是  nums[i] 右侧小于nums[i] 的元素的数量.例子:给定 nums = [5, 2, 6, 1]5的右侧有2个更小的元素 (2 和 1).2的右侧仅有1个更小的元素 (1).6的右侧有1个更小的元素 (1).1的右侧有0个更小的元素.返回数组 [2, 1, 1, 0].详见:https://leetcode.com/problems/count-of-smal…

You are given an integer array nums and you have to return a new countsarray. 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] Explanation: To the…

https://leetcode.com/problems/count-of-smaller-numbers-after-self/ 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…

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: Given nums = [5, 2, 6, 1] To the right of 5 there are…

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: Given nums = [5, 2, 6, 1] To the right of 5 there are…

原题链接在这里:https://leetcode.com/problems/count-of-smaller-numbers-after-self/ 题目: 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 r…

这是我在研究leetcode的solution第一个解决算法时,自己做出的理解,并且为了大家能看懂,做出了详细的注释. 此算法算是剑指Offer36的升级版,都使用的归并算法,但是此处的算法,难度更高,理解起来更加费劲. /* * @Param res 保存逆变对数 * @Param index 保存数组下标索引值,排序数组下标值. * 此算法使用归并算法,最大差异就在于merge()方法的转变 * * */ public List<Integer> countSmaller(int[] nu…

思路: bit + 离散化. 实现: #include <bits/stdc++.h> using namespace std; class Solution { public: int sum(vector<int> & bit, int i) { ; while (i) { ans += bit[i]; i -= i & -i; } return ans; } void add(vector<int> & bit, int i, int x)…

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…

Fenwick Tree Input The first line of input contains two integers NN, QQ, where 1≤N≤50000001≤N≤5000000 is the length of the array and 0≤Q≤50000000≤Q≤5000000 is the number of operations. Then follow QQ lines giving the operations. There are two types o…

Binary Indexed Tree 主要是为了存储数组前缀或或后缀和,以便计算任意一段的和.其优势在于可以常数时间处理更新(如果不需要更新直接用一个数组存储所有前缀/后缀和即可).空间复杂度O(n). 其中每个元素,存储的是数组中一段(起始元素看作为1而非0)的和: 假设这个元素下标为i,找到i的最低位1,从最低位1开始的低部表示的是长度,去除最低位1剩下的部分加上1表示的是起始位置,例如: 8二进制表示为100 最低位1也是最高位,从1开始的低部也即100本身,因此长度为8. 去除1以后,…

Description 现在请求你维护一个数列,要求提供以下两种操作: 1. 查询操作.语法:Q L 功能:查询当前数列中末尾L个数中的最大的数,并输出这个数的值.限制:L不超过当前数列的长度. 2. 插入操作.语法:A n 功能:将n加上t,其中t是最近一次查询操作的答案(如果还未执行过查询操作,则t=0),并将所得结果对一个固定的常数D取模,将所得答案插入到数列的 末尾.限制:n是非负整数并且在长整范围内.注意:初始时数列是空的,没有一个数. Input 第一行两个整数,M和D,其中M表示操…

Motivation: Given a 1D array of n elements. [2, 5, -1, 3, 6] range sum query: what's the sum from 2nd element to 4th element query(2, 4)? 5 + (-1) + 3 = 7 Native implementation: O(n) per query. Use DP to pre-compute the prefix sums in O(n), [2, 5, -1…

Fenwick Tree, (also known as Binary Indexed Tree,二叉索引树), is a high-performance data structure to calculate the sum of elements from the beginning to the indexed in a array. It needs three functions and an array: Array sum; It stores the data of Fenwi…

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

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…

Warning: input could be > 10000... Solution by segment tree: struct Node { Node(), left(nullptr), right(nullptr) {}; int start; int end; int cnt; // Node *left; Node *right; }; class Solution { Node *pRoot; void update(int v) { Node *p = pRoot; while…

[题目描述] 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…

2018-03-25 17:29:29 树状数组是一个比较小众的数据结构,主要应用领域是快速的对mutable array进行区间求和. 对于一般的一维情况下的区间和问题,一般有以下两种解法: 1)DP 预处理:建立长度为n的数组,每个结点i保存前i个数的和,时间复杂度O(n). 查询:直接从数组中取两个段相减,时间复杂度O(1). 更新:这种方法比较适用与immutable数组,对于mutable数组的更新需要重新建立表,所以时间复杂度为O(n). 2)树状数组 BIT 预处理:建立树状数组,…