49. Group Anagrams
这道题是242Valid Anagram这道题目的升级版，242是判断两个字符串是不是Anagram，现在是将多个Anagram字符串归类到一起。
class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
unordered_map<string, vector<string>> mp;
vector<vector<string>> res;
f......
滑动窗口经典题目，这里使用deque,双端队列中保存的是下标,保存值和下标均可以，依赖个人习惯。
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
deque<int> dq;
vector<int> ans;
for(int i = 0;i < nums.size();i++)
{
if(!dq.empty() && dq.front() == i - k)
dq.pop_front();
whi......
703. Kth Largest Element in a Stream
class KthLargest {
public:
KthLargest(int k, vector<int>& nums) {
size = k;
for(int i = 0;i < nums.size();i++)
{
pq.push(nums[i]);
if(pq.size() > size) pq.pop();
}
}
int add(int val) {
pq.push(val);
if(pq.size() > size)
pq.pop();
return pq.......
1021. Remove Outermost Parentheses
这道题的难点不在于编写程序，这么讲有点意思，leetcode的题目难点不在编程，这道题看半天没反应过来是什么意思，无奈打开力扣(leetcode中文版)，看了题目描述才明白题目的意思。简单讲就是去掉最外层的括号，这里先过滤了最左边的(，根据opened的值也过滤对应的).
class Solution {
public:
string removeOuterParentheses(string S) {
string res;
int opened = 0;
for (char c : S) {
if (c == '......
108. Convert Sorted Array to Binary Search Tree
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* ......
328. Odd Even Linked List
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solu......
1304. Find N Unique Integers Sum up to Zero
常规解法：
class Solution {
public:
vector<int> sumZero(int n) {
vector<int> res(n,0);
for(int i = 0;i < (n &~1);i+=2)
{
res[i] = i + 1,res[i + 1] = -(i + 1);
}
return res;
}
};
这里的n&~1作用是，如果n是奇数则执行n-1,如果n是偶数则不变。
1464. Maximum Product of Two Elements in an Array
无脑常规暴力破解方法：
class Solution {
public:
int maxProduct(vector<int>& nums) {
int n = nums.size();
int maximum = INT_MIN;
for(int i = 0;i < n;i++)
{
for(int j = i + 1;j < n;j++)
{
maximum = max(maximum,(nums[i] - 1) * (nums[j] - 1));
}......
1323. Maximum 69 Number
常规解法，略显冗余
class Solution {
public:
int maximum69Number (int num) {
stack<int> stack;
int nums = 0,changed = 0;
while(num)
{
stack.push(num % 10);
num /= 10;
}
while(!stack.empty())
{
if(stack.top() == 6 && !changed && (changed = 1))
nums = nums * 10......
1603. Design Parking System
常规解法：
class ParkingSystem {
public:
ParkingSystem(int big, int medium, int small) {
this->big = big;
this->medium = medium;
this->small = small;
}
bool addCar(int carType) {
switch(carType)
{
case 1:
if(big > 0){big--;return true;}else return false;break;......