chocoluffy

stringstream ss; while(getline(ss, symbol, ',')) {...}; // get every part split by seperator. stoi(char); // into int.

chocoluffy

基本上就是说不希望caller改变return的值，因此返回一个reference。

例如： const unordered_set<string> & dict

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https://stackoverflow.com/questions/8610350/implications-of-using-an-ampersand-before-a-function-name-in-c

It returns a reference to the private member.

chocoluffy

push: 直接将元素append进数据结构。只允许一个argument。对于stack<pair<int, char>>的情况，应该在argument处使用constructor: stack.push(std::make_pair(3, 'a'))，以保证只有一个argument。 emplace: 将元素放进数据结构，同时call constructor，允许多个argument，相当于简化了上述必须在argument处所做的处理: stack.emplace(3, 'a') 即可。

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https://stackoverflow.com/questions/4303513/push-back-vs-emplace-back

In addition to what visitor said :

The function void emplace_back(Type&& _Val) provided by MSCV10 is non conforming and redundant, because as you noted it is strictly equivalent to push_back(Type&& _Val).

chocoluffy

// https://www.geeksforgeeks.org/stack-data-structure-introduction-program/

// usually implemented with array or linked list implementation. 
// Or direcly using STL library: `#include <stack>`.

// following is an example of re-writing stack class, with extra function `max()`, returns the current max element of the stack.
#include <stack>
template <typename T>

class Stack {

chocoluffy

// Then sugguests that using only O(1) space complexity, meaning only `swap` is recommended!
// 很多关于linked list的技巧基本都和two pointer同时iterate有关。
// 1. using two pointer technique(slower and fast pointer) to find the middle point of the list.
// Q: check if a linked list is palindrome: two pointer iterate elements one by one and check if equal.
// Q: zip a linked list: interleaving the first and second halves of the linked list. 

chocoluffy

struct Node{
  int data;
  Node* next;
}

chocoluffy

node* reverse(node* L) {
  node* prev = NULL, curr = L;
  node* tmp = NULL;
  while(curr) {
    tmp = curr.next;
    curr.next = prev;
    prev = curr;
    curr = tmp;
  }
  return prev;

chocoluffy

// Q: given a list <l0, l1, l2, ...>, modify it into <l0, l2, l4,..., l1, l3...> a even-odd merged list.
// 多指针同时遍历的技巧。prev的使用。以及corner case针对even\odd的讨论。一个技巧是，根据已有的变量是否会seg fault来判断需要新增的case，最常见的遗漏就是对是否为null的讨论（针对linked list是否为空等等）。
node* even_odd_merge(node* L) {
  node* prev_even = NULL;
  node* even = L, odd = L.next;
  node* odd_start = L.next;
  while(even && odd) {
    prev_even = even;
    even.next = odd.next;
    even = even.next;