leetcode 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[i].

Example:

Given nums = [5, 2, 6, 1]To the right of 5 there are 2 smaller elements (2 and 1).To the right of 2 there is only 1 smaller element (1).To the right of 6 there is 1 smaller element (1).To the right of 1 there is 0 smaller element.

Return the array [2, 1, 1, 0].

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题目地址 leetcode Count of Smaller Numbers After Self

题意：

给定nums数组，求数组中每个元素i的右边比其小的数

思路：

简单的说就是求逆序数。

1. 使用逆序数有经典的解法为合并排序。
2. 用Fenwick树  关于Fenwick 树介绍 Binary indexed tree (Fenwick tree) 
   • 简单说就是看当前数在nums中排第几，然后对小于它的数求个数和

合并排序

struct Node {int val;int index;int cnt;Node(int val, int index) : val(val), index(index), cnt(0) {}bool operator <= (const Node &node2)const {return val <= node2.val;}};class Solution {public:void combine(vector<Node> &nums, int Lpos, int Lend, int Rend, vector<Node> &temp) {int Rpos = Lend + 1;int Tpos = Lpos;int n = Rend - Lpos + 1;int t = Rpos;while (Lpos <= Lend && Rpos <= Rend) {if (nums[Lpos] <= nums[Rpos]) {temp[Tpos] = nums[Lpos];temp[Tpos].cnt += Rpos - t ;Tpos++; Lpos++;}else {temp[Tpos++] = nums[Rpos++];}}while (Lpos <= Lend) {temp[Tpos] = nums[Lpos];temp[Tpos].cnt += Rpos - t;Tpos++; Lpos++;}while (Rpos <= Rend) temp[Tpos++] = nums[Rpos++];for (int i = 0; i< n; i++, Rend--) nums[Rend] = temp[Rend];}void merge_sort(vector<Node> & nums, int L, int R, vector<Node> &temp) {if (L < R) {int m = (L + R) >> 1;merge_sort(nums, L, m, temp);merge_sort(nums, m + 1, R, temp);combine(nums, L, m, R, temp);}}vector<int> countSmaller(vector<int>& nums) {vector<Node> mynums;vector<Node> temp(nums.size(), Node(0, 0));for (int i = 0; i < nums.size(); i++) mynums.push_back(Node(nums[i], i));vector<int> ans(nums.size(), 0);merge_sort(mynums, 0, nums.size() - 1, temp);for (int i = 0; i < nums.size(); i++) ans[mynums[i].index] = mynums[i].cnt;return ans;}};

Fenwick 树

class FenwickTree {vector<int> sum_array;int n;inline int lowbit(int x) {return x & -x;}public:FenwickTree(int n) :n(n), sum_array(n + 1, 0) {}void add(int x, int val) {while (x <= n) {sum_array[x] += val;x += lowbit(x);}}int sum(int x) {int res = 0;while (x > 0) {res += sum_array[x];x -= lowbit(x);}return res;}};class Solution {public:vector<int> countSmaller(vector<int>& nums) {vector<int> temp_num = nums;sort(temp_num.begin(), temp_num.end());unique(temp_num.begin(), temp_num.end());unordered_map<int,int> dic;for (int i = 0; i < temp_num.size(); i++) dic[temp_num[i]] = i + 1;FenwickTree tree(nums.size());vector<int> ans(nums.size(),0);for (int i = nums.size() - 1; i >= 0; i--) {ans[i] = tree.sum(dic[nums[i]] - 1);tree.add(dic[nums[i]],1);}return ans;}};

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