Maximum value of sliding window (STL application + Sword-point offer)

Maximum value of sliding window

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The title describes the size of the given array and the sliding window, and finds the maximum value of the values in all the sliding windows. For example, if you enter the array {2,3,4,2,6,2,5,1} and the size of the sliding window 3, there are 6 sliding windows, their maximum value is {4,4,6,6,6,5}, and the sliding window for the array {2,3,4,2,6,2,5,1} has the following 6: {[ 2,3,4],2,6,2,5,1}, {2,[3,4,2],6,2,5,1}, {2,3,[4,2,6],2,5,1}, {2,3,4,[2,6,2],5,1}, {2,3,4,2,[6,2,5],1}, {2,3,4,2,6,[ 2,5,1]}.Ideas:

With a double-ended queue, the first position of the queue holds the maximum value of the current window, when the window slides once
1. Determine if the current maximum value expires
2. The newly added value is compared from the tail of the team, throwing away all the values that are smaller than him.
monotone queue, O (n)

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The/** topic describes the size of the given array and sliding window, and finds the maximum value of the values in all the sliding windows. For example, if you enter the array {2,3,4,2,6,2,5,1} and the size of the sliding window 3, there are 6 sliding windows, their maximum value is {4,4,6,6,6,5}, and the sliding window for the array {2,3,4,2,6,2,5,1} has the following 6: {[ 2,3,4],2,6,2,5,1}, {2,[3,4,2],6,2,5,1}, {2,3,[4,2,6],2,5,1}, {2,3,4,[2,6,2],5,1}, {2,3,4,2,[6,2,5],1}, {2,3,4,2,6,[ 2,5,1]}.                                                    With a double-ended queue, the first position of the queue holds the maximum value of the current window, and when the window slides once 1. Determines whether the current maximum value expires 2. The new added value is compared from the end of the team, throwing away all the values that are smaller than him. Monotone queue, O (n) */#include <cstdio> #include <vector> #include <deque> #include <iostream> #include <algorithm>using namespace Std;class Solution {public:vector<int> maxinwindows (const vector<int>& Amp        num, unsigned int size) {vector<int> ans;        if (num.size () <size | | num.size () ==0 | | size<=0) return ans;  deque<int> tmp;        Double-ended queue int len=size,mmax=-1,mj=-1;                for (int i=0;i<len;i++) {if (Num[i]>mmax) {mmax=num[i];            Mj=i; }} for (int i=mj;i<len;i++) Tmp.push_back (i); All subscripts after the maximum value in Len should be placed in the container. Instead of just//Tmp.push_back (MJ);//The subscript of the largest value in the previous Len is placed in Ans.push_back (Mmax);//Place the maximum value of the former Len in the for (int I=LEN;I&LT;NUM.S            Ize (); i++) {//Deque<int>::iterator iter=tmp.begin ();//while (Iter!=tmp.end ())//            {//cout<<*iter++<< "";//}cout<<endl;            int t=num[i];  int K=tmp.front ();            The maximum value of the subscript if (i-k >= len)//Remove the already outdated maximum value!!!                   {while (i-k >= len) {tmp.pop_front ();                    if (!tmp.empty ()) K=tmp.front ();               else {tmp.push_back (i); break;}   }} int J=tmp.back ();                The last value of the subscript//Add a new number, and overwrite the previous can overwrite the value of the smaller than their own if (I!=j && num[i]>=num[j]) { while (Num[i]>=num[j]) {tmp.pop_back ();                   if (!tmp.empty ()) J=tmp.back ();                else break;            } tmp.push_back (i);                        } else Tmp.push_back (i);            K=tmp.front ();        Ans.push_back (Num[k]);    } return ans;    }};int Main () {vector<int> arr; unsigned int len=3;    /* Arr.push_back (2);    Arr.push_back (3);    Arr.push_back (4);    Arr.push_back (4);    Arr.push_back (6);    Arr.push_back (2);    Arr.push_back (3);    Arr.push_back (2);    Arr.push_back (5);    Arr.push_back (1); */Arr.push_back (16);    Arr.push_back (14);    Arr.push_back (12);    Arr.push_back (10);    Arr.push_back (8);    Arr.push_back (6);    Arr.push_back (4);    Arr.push_back (2);/* Arr.push_back (2);    Arr.push_back (3);    Arr.push_back (4);    Arr.push_back (2);    Arr.push_back (6);    Arr.push_back (2);    Arr.push_back (5);    Arr.push_back (1); */solution so;    Arr=so.maxinwindows (Arr,len); Vector<int>::itErator Iter=arr.begin ();    while (Iter!=arr.end ()) {cout<<*iter++<< "";    }cout<<endl; return 0;}

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Maximum value of sliding window (STL application + Sword-point offer)