if (condition?) then TODO() else if (condition?) then TODO() else TODO() end if
数组管理:
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arr = new array() // set a new array arr = [..., ..., ...] //set values for the array arr[...] //returns the value of a certain index arr.Length() //returns the length of an array(for some reason this could not run on the above website)
集合管理:
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coll = new collection() //set a new collection coll.addItem(certain things) //add something to the collection coll.getNext() //return the next value coll.resetNext() //go back to the start of the collection coll.hasNext() //returns a boolean value that indicates whether it has next item or not coll.isEmpty() //returns a boolean value that indicates whether the collection is empty or not
栈管理:
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// One thing to be remember, first in last out stack = new stack() //set a new stack stack.push(certain things) //push an item into the stack stack.pop() //returns the last item of the stack stack.isEmpty() //returns a boolean value that indicates whether the stack is empty or not
队列管理:
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// Another thing to remember, first in first out queue = new queue() //set a new queue queue.enqueue(certain things) //put an item to the end of the queue queue.dequeue() //delete and return the front item of the queue queue.isEmpty() //returns a boolean value that indicates whether the queue is empty or not
流程图画法
Then use arrows to link the blocks.
简单代码的伪代码实现
The following codes are all based on arrays. Searching/Sorting other form of data set need to do simple modifications to the code.
Sequential Search
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arr = [5, 6, 3, 4, 2, 7, 8, 1]
loop i from 0 to 7 if (arr[i] == 2) then index = i end if end loop
loop i from 0 to 6 min_idx = i loop j from i+1 to 7 if arr[j] < arr[min_idx] then min_idx = j end if end loop if NOT(min_idx == i) then TEMP = arr[min_idx] arr[min_idx] = arr[i] arr[i] = TEMP end if end loop //Sort the array first since binary search is only applicable in ordered series
low = 0 high = 7 found = -1 loop while found = -1 AND low <= high mid = div(low + high, 2) if (arr[mid] = 8) then found = mid else if (arr[mid] < 8) then low = mid + 1 else high = mid - 1 end if end loop
if found >= 0 then output "The index of number 8 is: ", found else output "Number was not found" end if
Bubble Sort
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arr = [5, 6, 3, 4, 2, 7, 8, 1]
loop i from 0 to 6 loop j from 0 to 6-i if arr[j] > arr[j+1] then TEMP = arr[j+1] arr[j+1] = arr[j] arr[j] = TEMP end if end loop end loop loop i from 0 to 7 output arr[i] end loop
Selection Sort
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arr = [5, 6, 3, 4, 2, 7, 8, 1] loop i from 0 to 6 min_idx = i loop j from i+1 to 7 if arr[j] < arr[min_idx] then min_idx = j end if end loop if NOT(min_idx == i) then TEMP = arr[min_idx] arr[min_idx] = arr[i] arr[i] = TEMP end if end loop