C++學習筆記（一）——leetcode記錄
阿新 • • 發佈：2018-12-19
C++學習筆記（一）——leetcode記錄
- 944. Delete Columns to Make Sorted [Easy]
- 852. Peak Index in a Mountain Array [Easy]
- 942. DI String Match [Easy]
- 922. Sort Array By Parity II [Easy]
- 933. Number of Recent Calls [Easy]
- 883. Projection Area of 3D Shapes [Easy]
- 590. N-ary Tree Postorder Traversal [Easy]
- 167. Two Sum II - Input array is sorted [Easy]
- 860. Lemonade Change [Easy]
- 697. Degree of an Array [Easy]
- 300. Longest Increasing Subsequence [Medium]
- 476. Number Complement [Easy]
- 268. Missing Number [Easy]
- 830. Positions of Large Groups [Easy]
- 876. Middle of the Linked List [Easy]
- 121. Best Time to Buy and Sell Stock [Easy]
- 628. Maximum Product of Three Numbers [Easy]
- 746. Min Cost Climbing Stairs [Easy]
Leetcode
最近學cpp，就寫題練練手，個人學習筆記，如存在問題，歡迎指出
944. Delete Columns to Make Sorted [Easy]
944. Delete Columns to Make Sorted [Easy]
class Solution {
public 852. Peak Index in a Mountain Array [Easy]
852. Peak Index in a Mountain Array [Easy]
class Solution {
public:
int peakIndexInMountainArray(vector<int>& A) {
int index = 0;
for(int i = 1; i < A.size(); i++){
if(A[i]>A[i-1]) index = i;
}
return index;
}
};
942. DI String Match [Easy]
class Solution {
public:
vector<int> diStringMatch(string S) {
vector<int> A;
for (int l = 0, h = S.size(), i = 0; i <= S.size(); i++) {
A.push_back(S[i] == 'I' ? l++ : h--);
}
return A;
}
};
922. Sort Array By Parity II [Easy]
922. Sort Array By Parity II [Easy]
class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& A) {
vector<int> odd, even;
for(int i = 0; i < A.size(); i++) {
if(A[i]%2 == 0) even.push_back(A[i]);
else odd.push_back(A[i]);
}
for(int i = 0; i < A.size()-1; i+=2) {
A[i] = even[i/2];
A[i+1] = odd[i/2];
}
return A;
}
};
933. Number of Recent Calls [Easy]
933. Number of Recent Calls [Easy]
題目大意:
找出最近的3000毫秒內有多少個呼叫請求。每個呼叫請求是ping(t)函式，其中t是請求的時間，可以保證每次ping的引數t是大於前面的。
reference
queue用法參考
class RecentCounter {
public:
queue<int> q;
RecentCounter() {
}
int ping(int t) {
while(!q.empty() && t-q.front()>3000) q.pop();
q.push(t);
return q.size();
}
};
/**
* Your RecentCounter object will be instantiated and called as such:
* RecentCounter* obj = new RecentCounter();
* int param_1 = obj->ping(t);
*/
883. Projection Area of 3D Shapes [Easy]
883. Projection Area of 3D Shapes [Easy]
class Solution {
public:
int projectionArea(vector<vector<int>>& grid) {
int res = 0, n = grid.size(), m = grid[0].size();
for (int i = 0, v = 0; i < n; ++i, res += v, v = 0)
for (int j = 0; j < m; ++j)
v = max(v, grid[i][j]);
for (int j = 0, v = 0; j < m; ++j, res += v, v = 0)
for (int i = 0; i < n; ++i)
v = max(v, grid[i][j]);
for (int i = 0; i < n; ++i)
for (int j = 0; j < m; ++j)
if (grid[i][j]) res++;
return res;
}
};
int projectionArea(const vector<vector<int>>& grid) {
int res = 0, n = grid.size(), x, y;
for (int i = 0; i < n; ++i) {
x = 0, y = 0;
for (int j = 0; j < n; ++j) {
x = max(x, grid[i][j]);
y = max(y, grid[j][i]);
if (grid[i][j]) ++res;
}
res += x + y;
}
return res;
}
590. N-ary Tree Postorder Traversal [Easy]
590. N-ary Tree Postorder Traversal [Easy]
/*
// Definition for a Node.
class Node {
public:
int val;
vector<Node*> children;
Node() {}
Node(int _val, vector<Node*> _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public:
vector<int> postorder(Node* root) {
if(root==NULL) return {};
vector<int> res;
stack<Node*> stk;
stk.push(root);
while(!stk.empty())
{
Node* temp = stk.top();
stk.pop();
for(int i = 0; i < temp->children.size(); i++) stk.push(temp->children[i]);
res.push_back(temp->val);
}
reverse(res.begin(), res.end());
return res;
}
};
stack的定義：C++stack(堆疊）是一個容器的改編，它實現了一個先進後出的資料結構（FILO），不允許被遍歷，沒有迭代器。 使用該容器時需要包含#include標頭檔案.
**top()**是取棧頂元素
**pop()**是彈出棧頂元素
167. Two Sum II - Input array is sorted [Easy]
167. Two Sum II - Input array is sorted [Easy]
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
vector<int> res;
int end = numbers.size()-1;
int i = 0;
int j = end;
for(int i = 0; i < j; i++){
for(int j = end; j > i; j--){
if(numbers[i] + numbers[j] == target) {
res.push_back(i+1);
res.push_back(j+1);
return res;
}else if(numbers[i] + numbers[j] < target){
end = j;
break;
}else continue;
}
}
}
};
其他簡單方法：
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int start = 0, end = numbers.size()-1;
while(start < end) {
if(numbers[start] + numbers[end] == target) {
return {start+1, end+1};
}
if(numbers[start] + numbers[end] > target) {
end--;
}
else {
start++;
}
}
}
};
vector<int> twoSum(vector<int>& numbers, int target) {
int l = 0;
int r = numbers.size() -1;
while(l < r){
if(numbers[l] + numbers[r] == target){
vector<int> res{l+1,r+1};
return res;
}
else if(numbers[l] + numbers[r] > target){
r--;
}
else{
l++;
}
}
}
860. Lemonade Change [Easy]
class Solution {
public:
bool lemonadeChange(vector<int>& bills) {
int five = 0, ten = 0;
for(int i = 0; i < bills.size(); i++){
if(bills[i] == 5) five++;
else if(bills[i] == 10){
if(five == 0) return 0;
five--;
ten++;
}else{
if(five>0 && ten>0){
ten--;
five--;
}else if(five > 2) five -= 3;
else return 0;
}
}
return 1;
}
};
697. Degree of an Array [Easy]
697. Degree of an Array [Easy]
class Solution {
public:
int findShortestSubArray(vector<int>& nums) {
unordered_map<int,vector<int>> mp;
for(int i=0;i<nums.size();i++) mp[nums[i]].push_back(i);
int degree=0;
for(auto it=mp.begin();it!=mp.end();it++) degree=max(degree,int(it->second.size()));
int shortest=nums.size();
for(auto it=mp.begin();it!=mp.end();it++){
if(it->second.size()==degree){
shortest=min(shortest,it->second.back()-it->second[0]+1);
}
}
return shortest;
}
};
unordered_map
template < class Key, // unordered_map::key_type
class T, // unordered_map::mapped_type
class Hash = hash<Key>, // unordered_map::hasher
class Pred = equal_to<Key>, // unordered_map::key_equal
class Alloc = allocator< pair<const Key,T> > // unordered_map::allocator_type
> class unordered_map;
In the reference for the unordered_map member functions, these same names (Key, T, Hash, Pred and Alloc) are assumed for the template parameters.
Iterators to elements of unordered_map containers access to both the key and the mapped value. For this, the class defines what is called its value_type, which is a pair class with its first value corresponding to the const version of the key type (template parameter Key) and its second value corresponding to the mapped value (template parameter T):
typedef pair<const Key, T> value_type;
Iterators of a unordered_map container point to elements of this value_type. Thus, for an iterator called it that points to an element of a map, its key and mapped value can be accessed respectively with:
unordered_map<Key,T>::iterator it;
(*it).first; // the key value (of type Key)
(*it).second; // the mapped value (of type T)
(*it); // the "element value" (of type pair<const Key,T>)
Naturally, any other direct access operator, such as -> or [] can be used, for example:
it->first; // same as (*it).first (the key value)
it->second; // same as (*it).second (the mapped value)
300. Longest Increasing Subsequence [Medium]
300. Longest Increasing Subsequence [Medium]
class Solution {
public:
int lengthOfLIS(vector<int>& nums) {
if(nums.size() == 0) return 0;
vector<int> length(nums.size());
int longest = 1;
length[0] = 1;
for(int i = 1; i < nums.size(); i++){
length[i] = 1;
for(int j = 0; j < i; j++){
if(nums[j] < nums[i]) length[i] = max(length<