LeetCode日记20200218
LeetCode日记2020.2.18
文章目录
- LeetCode日记2020.2.18
- 723 我的日程安排表(hard)
- 986 区间列表的交集(mid)
- 442 数组中的重复数据(mid)
- 64 最小路径和(mid)
- 238 除自身以外数组的乘积(mid)
- 106 从中序与后序遍历序列构造二叉树(mid)
- 48 旋转图像(mid)
5数组
+ 1随机
mid + 1随机
hard
723 我的日程安排表(hard)
自己的较慢速解法，以后要用set时问问自己：是否会有重复数据？
class MyCalendarThree {
public:
MyCalendarThree() {
}
int book(int start, int end) {
calen.insert(make_pair(start, end));
int res = 0;
priority_queue<int, vector<int>, greater<int>> pq;
for(const auto& c: calen)
{
while(!pq.empty() && pq.top() <= c.first)
{
pq.pop();
}
pq.push(c.second);
res = max(res, int(pq.size()));
}
return res;
}
multiset<pair<int, int>> calen;
};
/**
* Your MyCalendarThree object will be instantiated and called as such:
* MyCalendarThree* obj = new MyCalendarThree();
* int param_1 = obj->book(start,end);
*/
优化后快速解法：
class MyCalendarThree {
public:
MyCalendarThree() {
}
int book(int start, int end) {
if(calen.find(start) != calen.end())
++calen[start];
else
calen[start] = 1;
if(calen.find(end) != calen.end())
--calen[end];
else
calen[end] = -1;
int res = 0;
int k = 0;
for(const auto& c: calen)
{
k += c.second;
if(k>res)
res = k;
}
return res;
}
map<int, int> calen;
};
/**
* Your MyCalendarThree object will be instantiated and called as such:
* MyCalendarThree* obj = new MyCalendarThree();
* int param_1 = obj->book(start,end);
*/
986 区间列表的交集(mid)
注意利用每个列表内区间排序且互不相交的性质。
class Solution {
public:
vector<vector<int>> intervalIntersection(vector<vector<int>>& A, vector<vector<int>>& B) {
vector<vector<int>> res;
int i=0, j=0;
while(i<A.size()&&j<B.size())
{
if(A[i][0]<=B[j][1] && A[i][1]>=B[j][0])
res.push_back(vector<int>{max(A[i][0], B[j][0]), min(A[i][1], B[j][1])});
if(A[i][1] == B[j][1])
{
++i;
++j;
}
else if(A[i][1]<B[j][1])
++i;
else
++j;
}
return res;
}
};
442 数组中的重复数据(mid)
class Solution {
public:
vector<int> findDuplicates(vector<int>& nums) {
vector<int> res;
for(int i=0;i<nums.size();++i)
{
int num = abs(nums[i]);
if(nums[num-1]<0)
res.push_back(num);
nums[num-1] = -nums[num-1];
}
return res;
}
};
64 最小路径和(mid)
class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
int m = grid.size(), n = grid.front().size();
vector<int> dp(n+1, 0);
for(int i=n-1;i>=0;--i)
dp[i] = dp[i+1] + grid[m-1][i];
dp[n] = numeric_limits<int>::max();
for(int i=m-2;i>=0;--i)
{
for(int j=n-1;j>=0;--j)
{
dp[j] = grid[i][j] + min(dp[j], dp[j+1]);
}
}
return dp[0];
}
};
238 除自身以外数组的乘积(mid)
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int n = nums.size();
vector<int> res(n, 1);
int prod = 1;
for(int i=0;i<n;++i)
{
res[i] *= prod;
prod *= nums[i];
}
prod = 1;
for(int i=n-1;i>=0;--i)
{
res[i] *= prod;
prod *= nums[i];
}
return res;
}
};
106 从中序与后序遍历序列构造二叉树(mid)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
using Iter = vector<int>::const_iterator;
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
reverse(postorder.begin(), postorder.end());
return buildT(inorder.cbegin(), inorder.cend(), postorder.cbegin(), postorder.cend());
}
TreeNode* buildT(Iter inBeg, Iter inEnd, Iter pBeg, Iter pEnd)
{
if(pBeg == pEnd)
return NULL;
int val = *pBeg;
TreeNode* root = new TreeNode(val);
auto pos = find(inBeg, inEnd, val);
int dis = inEnd - pos - 1;
auto lpEnd = pBeg + 1;
advance(lpEnd, dis);
root->right = buildT(pos+1, inEnd, pBeg + 1, lpEnd);
root->left = buildT(inBeg, pos, lpEnd, pEnd);
return root;
}
};
48 旋转图像(mid)
class Solution {
public:
void rotate(vector<vector<int>>& matrix) {
int n=matrix.size();
for(int i=0;i<n;++i)
{
for(int j=i;j<n;++j)
{
std::swap(matrix[i][j], matrix[j][i]);
}
}
for(int i=0;i<n;++i)
{
for(int j=0;j<n/2;++j)
{
std::swap(matrix[i][j], matrix[i][n-1-j]);
}
}
}
};