"Deque" sliding window, double-ended queue solves array problem

C + + Handbook of Deque

Owning header file <deque>

Common operations:

Back () returns the trailing element;

Front () returns the head element;

Push_back () tail insertion element;

POP_BAKC () tail delete element;

Push_front () head insertion element;

Pop_front () head delete element;

Question 1: Find the maximum value of the sliding window ("The sword refers to the offer surface question 65")

Description: Given an array and the size of the sliding window, find the maximum value in all the sliding windows.

Example: The array {2, 3, 4, 2, 6, 2, 5} has a window size of 3, a total of 7-3+1=5 sliding windows, returns an array of the maximum values of these windows {4, 4, 6, 6, 6}

Method: Generate a double-ended queue deque, which stores the subscript that may be the maximum value of a sliding window in an array.

vector<int> Getmax (Constvector<int> &arr,intsize) {Vector<int>Res; if(arr.size () = =0|| Size <=0){        returnRes; } deque<int>index; inti =0;  while(I <arr.size ()) {         while(!index.empty () && arr[index.back ()] <=Arr[i])        {Index.pop_back ();        } index.push_back (i); if(!index.empty () && index.front () = = (I-size)) {Index.pop_front ();//does not belong to the current window        }        if(I >= Size-1) {Res.push_back (Arr[index.front ())); } I++; }    returnRes;}

Question 2: Given an array, find out the difference between the maximum and minimum values is less than the number of sub-arrays equal to the given number Num. (New Ket left Chengyun "Programmer interview Code Guide")

Example: Given array {5, 2, 6, 3, 4, 1},num = 3, the eligible sub-arrays are {5} {5}, 2} {2} {6} {6, 3} {6}, 3, 4} {3} {3, 4} {3, 4, 1} {4} {} 4

Method: Generate two double-ended queue Qmax, Qmin, the first-level while loop I, satisfies the condition of I as the starting element of the sub-array, the second-level while loop, satisfies the condition of j as the tail element of the sub-array. Each time the inner while loop ends, the result res + = J-i;

Qmax maintains the subscript of the maximum value of the window sub-array in the original array;

Qmin maintains the subscript of the window sub-array minimum value in the original array.

The window changes dynamically, and the elements of Qmax and qmin are also changing.

1 intGetnum (intArr[],intLenintnum) {2     if(arr = NULL | | Len <1|| Num <0)3         return 0;4     intres =0;5deque<int>Qmax;6deque<int>qmin;7     inti =0;8     intj =0;9      while(I <Len) {Ten          while(J <Len) { One              while(!qmax.empty () && arr[qmax.back ()] <=Arr[j]) { A Qmax.pop_back (); -             } - Qmax.push_back (j); the              while(!qmin.empty () && arr[qmin.back ()] >=Arr[j]) { - Qmin.pop_back (); -             } - Qmin.push_back (j); +             if(Arr[qmax.front ()]-Arr[qmin.front ()] >num) { -                  Break; +             } AJ + +The//j has been getting bigger, not back. at         } -         if(Qmax.front () = =i) { - Qmax.pop_front ();//does not belong to a window that starts with i+1, pops up -         } -         if(Qmin.front () = =i) { - Qmin.pop_front ();//does not belong to a window that starts with i+1, pops up in         } -Res + = J-i; toi++the//i has been getting bigger, not back. +     } -     returnRes; the}

"Deque" sliding window, double-ended queue solves array problem