题目连接

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

Count of Smaller Numbers After Self

Description

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 $[2, 1, 1, 0]$.

线段树求逆序数，先离散化，再用线段树统计。。

class Solution {
private:
	typedef vector<int> vec;
public:
	vec countSmaller(vec& nums) {
		vec ans;
		int n = 0;
		if (!(n = nums.size())) return ans;
		ans.resize(n, 0);
		ret = new P[n + 10];
		for (int i = 1; i <= n; i++) ret[i] = P(nums[i - 1], i);
		sort(ret + 1, ret + n + 1);
		seg = new int[(n + 10) << 2];
		memset(seg, 0, sizeof(int)* ((n + 10) << 2));
		for (int i = 1; i <= n; i++) {
			int v = query(1, 1, n, ret[i].id, n);
			insert(1, 1, n, ret[i].id);
			ans[ret[i].id - 1] = v;
		}
		delete[]seg; delete[]ret;
		return ans;
	}
private:
	struct P {
		int v, id;
		P(int _i_ = 0, int _j_ = 0) :v(_i_), id(_j_) {}
		friend bool operator<(const P &a, const P &b) {
			return a.v == b.v ? a.id < b.id : a.v < b.v;
		}
	}*ret;
	int *seg;
	inline void insert(int root, int l, int r, int p) {
		if (p > r || p < l) return;
		if (p <= l && p >= r) { seg[root]++; return; }
		int mid = (l + r) >> 1;
		insert(root << 1, l, mid, p);
		insert(root << 1 | 1, mid + 1, r, p);
		seg[root] = seg[root << 1] + seg[root << 1 | 1];
	}
	inline int query(int root, int l, int r, int x, int y) {
		if (x > r || y < l) return 0;
		if (x <= l && y >= r) return seg[root];
		int ret = 0;
		int mid = (l + r) >> 1;
		ret += query(root<<1, l, mid, x, y);
		ret += query(root << 1 | 1, mid + 1, r, x, y);
		return ret;
	}
};