标签：
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 $nums[i]$.
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;
}
};
leetcode Count of Smaller Numbers After Self
标签：
原文地址：http://www.cnblogs.com/GadyPu/p/5024052.html
