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 &nums, int Lpos, int Lend, int Rend, vector &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 & nums, int L, int R, vector &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 countSmaller(vector& nums) {
		vector mynums;
		vector temp(nums.size(), Node(0, 0));
		for (int i = 0; i < nums.size(); i++) 
			mynums.push_back(Node(nums[i], i));
		
		vector 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 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 countSmaller(vector& nums) {
		vector temp_num = nums;
		sort(temp_num.begin(), temp_num.end());
		unique(temp_num.begin(), temp_num.end());
		unordered_map dic;
		for (int i = 0; i < temp_num.size(); i++) 
			dic[temp_num[i]] = i + 1;

		FenwickTree tree(nums.size());
		vector 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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