[LeetCode] Count of Smaller Numbers After Self 计算后面较小数字的个数

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].

// Binary Search
class Solution {
public:
vector<int> countSmaller(vector<int>& nums) {
vector<int> t, res(nums.size());
for (int i = nums.size() - 1; i >= 0; --i) {
int left = 0, right = t.size();
while (left < right) {
int mid = left + (right - left) / 2;
if (t[mid] >= nums[i]) right = mid;
else left = mid + 1;
}
res[i] = right;
t.insert(t.begin() + right, nums[i]);
}
return res;
}
};

// Insert Sort
class Solution {
public:
vector<int> countSmaller(vector<int>& nums) {
vector<int> t, res(nums.size());
for (int i = nums.size() - 1; i >= 0; --i) {
int d = distance(t.begin(), lower_bound(t.begin(), t.end(), nums[i]));
res[i] = d;
t.insert(t.begin() + d, nums[i]);
}
return res;
}
};

// Binary Search Tree
class Solution {
public:
struct Node {
int val, smaller;
Node *left, *right;
Node(int v, int s) : val(v), smaller(s), left(NULL), right(NULL) {}
};
int insert(Node *&root, int v) {
if (!root) return (root = new Node(v, 0)), 0;
if (root->val > v) return root->smaller++, insert(root->left, v);
else return insert(root->right, v) + root->smaller + (root->val < v ? 1 : 0);
}
vector<int> countSmaller(vector<int>& nums) {
vector<int> res(nums.size());
Node *root = NULL;
for (int i = nums.size() - 1; i >= 0; --i) {
res[i] = insert(root, nums[i]);
}
return res;
}
};

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