我不想刷题

(Updated: )

刷题还是得多反思复盘总结…开个post吧

先把之前写的一些奇奇怪怪的东西贴过来:

Read questions carefully.

Matrix

how to flatten a 2d matrix?

2d matrix <-> 1d array

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M[i][j] = M[n*i+j]
M[i] = M[i/n][i%n]

n: column of the matrix

search in a matrix

Some matrix have descending/non-descending row or columns.
In such cases, we can:

  • use binary search.
  • assign a cursor.
  • …

Two Pointers

Intervals - looking for boundaries

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int start = Math.max(1, 2);
int end = Math.min(1, 2);

if(start <= end)
interval

Sliding window

0,0 -> 0,1 … 0,size
for(right)
right go ahead, make window
then move left

Tree

postorderTraversal

postorderTraversal is opposite to preorderTraversal, so we can put preorderTraversal to a stack, then pop the postorderTraversal.

BST

sorted array => BST
middle element is the root node

backtrack

use a linkedlist as track or use arguments
like tree traversal, when we get the base case, add result then return.

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we add track(make decision)
recursion()
remove track(recall decision)

Strings / Chars

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int[ ] hash = new int[26/256];

think reversely sometimes helps.

  • iterating from the smallest? or largest?
  • if A->B is difficult, how about use B to verify A?

Greedy (??????)

Prefix Sum

curSum - prevSum = k (prev, cur)
curSum - k = prevSum

min
secondMin