Generating an array of maximum window values

1. Generating a maximum array of window valuesThere is an integer array of arr and a window of size w that slides from the leftmost edge of the array to the far right, and the window slides one position at a time to the right. For example, the array is [4,3,5,4,3,3,6,7] and the window size is 3 o'clock: [4 3 5] 4 3 3 6 7 The maximum value in the window is 54 [3 5 4] 3 3 6 7 The maximum value in the window is 54 3 [5 4 3] 3 6 7 The maximum value in the window is 54 3 5 [4 3 3] 6 7 window maximum value is 44 3 5 4 [3 3 6] 7 window maximum value is 64 3 5 4 3 [3 6 7] The maximum value in the window is 7 if the array length is n and the window size is W, the maximum value of the N-w+1 window is generated altogether. Please implement a function given an array of arr, window size W. Returns an array of length n-w+1 res,res[i] represents the maximum value for each window state. Taking the case as an example, the result should return [5,5,5,4,6,7]. 2. The maximum value minus the minimum number of sub-arrays that are less than or equal to NumGiven the array arr and integer num, returns how many sub-arrays meet the following conditions: Max (arr[i. J])-min (arr[i). J]) <= Nummax (arr[i). J]) represents a subarray of arr[i. J] The maximum value, min (arr[i). J]) represents a subarray of arr[i. The minimum value in J]. If the array length is n, implement a solution with a time complexity of O (n). The reference code is as follows: C + + Code # # requires C + + 11 Support # #

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#include<iostream> #include<cmath> #include<string.h> #include<vector> #include<deque> using namespacestd; #defineLENGTH (ARR) (sizeof(ARR)/sizeof(arr[0])) /** @brief * Generate a maximum array of window values * Implement a function, given an array of arr, window size W. Returns an array of length n-w+1 res * Res[i] Indicates the maximum value for each window state * * @param arr[] int * @param length INT-array Lengths * @param winlen INT-Window size * @return vector<int>-array of maximum values * */ vector<int> Slidingwindowmaxarray (intarr[],intlength,intWinlen) { deque<int> Maxqueue;//Double-ended queue, store subscript, corresponding elements descending order vector<int> ret;//Results for(inti =0; i < length; ++i) { /**< Queue Tail-out queue if current element >= Queue trailing element * / while(!maxqueue.empty () && arr[maxqueue.back ()] <= arr[i]) { Maxqueue.pop_back (); } Maxqueue.push_back (i); //Current element subscript enters queue tail if(maxqueue.front () = = I-winlen)///Current window range >w, the queue head out of queue { Maxqueue.pop_front (); } if(i >= Winlen-1) //initial possible window range <w         { Ret.push_back (Arr[maxqueue.front ()); } } returnret; } /** @brief * Maximum number of sub-arrays minus the minimum value less than or equal to num * Given an array of arr and integer num, returns how many sub-arrays meet the following conditions: * MAX (arr[i. J])-min (arr[i). J]) <= num * * @param arr[] int * @param length int * @param num INT * @return int * */ intAlllessnumsubarray (intarr[],intlength,intnum) { deque<int> Mindeque; deque<int> Maxdeque; intret =0; intLow =0; intHigh =0; while(Low < length) { while(High < length) { while(!mindeque.empty () && arr[mindeque.back ()] >= Arr[high]) { Mindeque.pop_back (); } Mindeque.push_back (high); while(!maxdeque.empty () && arr[maxdeque.back ()] <= Arr[high]) { Maxdeque.pop_back (); } Maxdeque.push_back (high); if(Arr[maxdeque.front ()]-Arr[mindeque.front ()] > num) Break; ++high; } if(mindeque.front () = = Low) Mindeque.pop_front (); if(maxdeque.front () = = Low) Maxdeque.pop_front (); RET + = high-low; ++low; } returnret; } intMain () { intarr[] = {4, 3, 5, 4, 3, 3, 6, 7}; for(Autoval:slidingwindowmaxarray arr., LENGTH (arr),3)) { cout << Val <<" "; } cout << Endl; Output Result "5 5 5 4 6 7" intarr1[] = {7, 9, 6, 1, 0, 7, 5, 4, 4, 4, 2, 0, 7, 1, 7, 2, 5, 3, 1, 9, 0, 8, 8, 9, 4, 2, 3, 6, 9, 8 }; cout << Alllessnumsubarray (arr1, LENGTH (arr1),5) << Endl; //79 return 0; }

Generating an array of maximum window values