Stack and queue----maximum minus the minimum number of sub-arrays equal to num

Maximum minus the minimum value less than the number of sub-arrays equal to num

　　

Given the array arr and integer num, total returns how many arrays meet the following conditions: Max (arr[i. J])-min (arr[i). J]) <=num. where Max (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 length of the array is n, the time complexity is required to be O (n).

Resolution

Using a double-ended queue, Qmax maintains the window sub-array arr[i. J] The maximum value of the updated structure, Qmin maintains the window sub-array arr[i. J] The minimum value of the updated structure, all subscript values are up to Qmax and Qmin once, out of Qmax and Qmin once, and the values of I and J are also increasing, never decreasing, so the time complexity is O (N).

The following two conclusions can be obtained through the analysis of the topic:

1) If the sub-array arr[i: J] satisfies the condition, that is, Max (arr[i. J])-min (arr[i). J]) <=num, then arr[k. L] (I<=K<=L<=J) must satisfy the condition, even if an array satisfies the condition, all its sub-arrays must satisfy the condition.

2) If the sub-array arr[i: J] does not satisfy the condition, that is, Max (arr[i. J])-min (arr[i). J]) >num, then arr[k. L] (k<=i<=j<=l) certainly does not satisfy the condition, even if an array does not satisfy the condition, all the array containing it must not satisfy the condition.

　　

 Packagecom.test;Importjava.util.LinkedList;/*** Created by DEMRYSTV. * * All subscript values are up to Qmax and Qmin once, out of Qmax and qmin once, the values of I and J are increasing, never decreasing, so the time complexity is O (N)*/ Public classGetnum { Public intGetnum (int[] arr,intnum) {        if(Arr.length==0 | | arr==NULL){            return0; }        //double-ended queue, Qmax maintains the window sub-array arr[i. J] The maximum value of the updated structure, Qmin maintains the window sub-array arr[i. J] The minimum value of the updated structure,//if an array satisfies the criteria, then the sub-arrays it contains must all satisfy the condition//If an array does not satisfy the condition, then all arrays containing it must not satisfy the conditionLinkedlist<integer> Qmax =NewLinkedlist<integer>(); LinkedList<Integer> qmin =NewLinkedlist<integer>(); inti = 0; intj = 0; intres = 0;  while(i<arr.length) {             while(j<arr.length) {                 while(!qmin.isempty () && arr[qmin.peeklast ()]>=Arr[j])                {qmin.polllast ();                } qmin.addlast (j);  while(!qmax.isempty () && arr[qmax.peeklast ()]<=Arr[j])                {qmax.polllast ();                } qmax.addlast (j); if(Arr[qmax.getfirst ()]-Arr[qmin.getfirst ()] >num) {                     Break; } J++; }            //double-ended queue the team head to the starting point, he bounced out            if(Qmin.peekfirst () = =i)            {Qmin.pollfirst (); }            if(Qmax.peekfirst () = =i)            {Qmax.pollfirst (); }            //the use of the conclusion is if arr[i. J] Meet the conditions, then arr[i. J-1] must meet the conditions, if arr[i. J] satisfies the condition, then contains the arr[i. J] Array must not satisfy the conditionRes + = J-i; I++; }        returnRes; }}

Stack and queue----maximum value minus the minimum number of sub-arrays equal to num