leetcode@ [315/215] Count of Smaller Numbers After Self / Kth Largest Element in an Array (BST)
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]
.
class Solution {
class TreeNode{
public: int val, rank;
TreeNode *l, *r;
TreeNode(int v): val(v), rank(), l(NULL), r(NULL) {}
};
public:
int getRank(TreeNode* root, int v) {
int rank = ;
while(true) {
if(v <= root->val) {
++root->rank;
if(root->l == NULL) {
root->l = new TreeNode(v);
break;
}
else root = root->l;
}
else{
rank += root->rank;
if(root->r == NULL) {
root->r = new TreeNode(v);
break;
}
else root = root->r;
}
}
return rank;
} vector<int> countSmaller(vector<int>& nums) {
vector<int> res; if(nums.size() == ) return res;
TreeNode* root = new TreeNode(nums[nums.size()-]); res.push_back();
for(int i=nums.size()-; i>=; --i) {
int rank = getRank(root, nums[i]);
res.push_back(rank);
} vector<int> rev_res;
for(vector<int>::reverse_iterator p = res.rbegin(); p!=res.rend(); ++p) rev_res.push_back(*p);
return rev_res;
}
};
https://leetcode.com/problems/kth-largest-element-in-an-array/
Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.
For example,
Given [3,2,1,5,6,4]
and k = 2, return 5.
Note:
You may assume k is always valid, 1 ≤ k ≤ array's length.
class Solution {
class TreeNode{
public: int val, rank;
TreeNode *l, *r;
TreeNode(int v): val(v), rank(), l(NULL), r(NULL) {}
}; public:
void addNode(TreeNode* root, int v) {
while(true) {
if(v <= root->val) {
++root->rank;
if(root->l == NULL) {
root->l = new TreeNode(v);
break;
}
else root = root->l;
}
else{
if(root->r == NULL) {
root->r = new TreeNode(v);
break;
}
else root = root->r;
}
}
} void dfs(TreeNode* root, int k, int& res) {
if(root->rank == k) {
res = root->val;
return;
} if(root->l) dfs(root->l, k, res);
if(root->r) dfs(root->r, k - root->rank, res);
} int findKthLargest(vector<int>& nums, int k) {
if(nums.size() == ) return nums[]; TreeNode *root = new TreeNode(nums[]);
for(int i=; i<nums.size(); ++i) {
addNode(root, nums[i]);
} int res = -;
dfs(root, nums.size() - k + , res);
return res;
}
};
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