最近开始低频做题,主要是时间太久了都忘了,而且我C++也不熟,写几道题可以熟悉一下C++。工作中hack用的挺多的,C++ 和 python 也在用但没用那么多。
这道题有多种解法。可以用priorityQueue来做,也可以用segmentTree来做, 也可以用binary index tree来做。
我先写个Segment Tree的解法
struct MyTreeNode {
int start;
int end;
int availableSlots;
MyTreeNode* left;
MyTreeNode* right;
MyTreeNode(int a, int b, int slots): start(a), end(b), availableSlots(slots), left(nullptr), right(nullptr) {}
};
class SegmentTree {
private:
MyTreeNode* root;
public:
SegmentTree(int start, int end): root(buildTree(start, end)) {
}
void fill(vector& people, vector>& ans) {
fill(people, ans, root, people[1] + 1);
}
private:
MyTreeNode* buildTree(int start, int end) {
if (start > end) return nullptr;
if (start == end) {
MyTreeNode* node = new MyTreeNode(start, end, 1);
return node;
}
int mid = start + (end - start) / 2;
MyTreeNode* node = new MyTreeNode(start, end, end - start + 1);
node->left = buildTree(start, mid);
node->right = buildTree(mid + 1, end);
return node;
}
void fill(vector& people, vector>& ans, MyTreeNode* node, int pos) {
if (node == nullptr) return;
node->availableSlots--;
if (node->start == node->end) {
ans[node->start] = vector {people[0], people[1]};
return;
}
int leftAvailableSlots = node->left->availableSlots;
if (leftAvailableSlots >= pos) fill(people, ans, node->left, pos);
else fill(people, ans, node->right, pos - leftAvailableSlots);
}
};
class Solution {
public:
vector> reconstructQueue(vector>& people) {
sort(begin(people), end(people), [](const auto& x, const auto& y) {
return x[0] != y[0] ? x[0] < y[0] : x[1] > y[1];
});
SegmentTree tree {0, (int) people.size() - 1};
vector> ans(people.size(), vector(2,0));
for(auto& person: people) {
tree.fill(person, ans);
}
return ans;
}
};
C++的几个知识点
- sort(begin(people), end(people),...)这个用法
- sort里面 lambda函数
- vector
> ans(people.size(), vector (2,0)); 初始化
工作中这么写会不会内存泄漏,在何处删掉那些指针 ?