Fortune left France advanced classes 4_2 accumulation and aim for the longest sub-array

【topic】

The length of a given array arr, AIM and an integer, in seeking arr, the cumulative sum is equal to the longest num subarray

example:

arr = { 7,3,2,1,1,7,7,7 } aim = 7

There are many sub-array cumulative sum is equal to 7, but the longest subarray is {3,2,1,1}, it returns to its length 4

[Solving]

Using the map, the number of the array from the first accumulation start, recording and accumulating a first position to a certain value

The initial map data storage <0, -1> as there is no accumulation of a number and is set to 0, the position of the digital array - the array position [i.e., non-traversing]

map <sum, index>, when the sum unchanged, map does not update! ! !

note:

We're looking for sum - aim = index == key first appeared in the map of the location, and records

For example:? 7 3 2 1 1 7 - 6 - 1 7 num == 7

The cumulative numbers of 0 7, sum = 7, sum - aim = 0, map the present position key = 0, the position of the earliest - 1,

DESCRIPTION able to form a sub-array, and so that the accumulation of AIM, the array length of 0 - (- 1) = 1

Then continue to traverse backwards when traversing to the fourth digit 1,

map记录为{ <0,-1><7,0><10,1><12,2><13,3><14,4> }，

At this sum is 14, sum - aim = 14 - 7 = 7, key = 7 present the map, the position of the first occurrence is 0,

DESCRIPTION able to form a sub-array, and so that the accumulation of AIM, the array length of 4 - (0) = 4

 

[Code]

　　

. 1  #pragma Once
 2 #include <the iostream>
 . 3 #include <Map>
 . 4 #include <Vector>
 . 5  
. 6  the using  namespace STD;
 . 7  
. 8  int mostLongSubArray (Vector < int > V, int AIM)
 . 9  {
 10      int RES = 0 ; // record the longest recording 
. 11      Map < int , int > m; // record accumulation and 
12 is      m [ 0 ] = - . 1 ; // beginning position of the sum is zero 
13 is     int SUM = 0 ;
 14      for ( int I = 0 ; I <v.size (); ++ I)
 15      {
 16          SUM = + V [I];
 . 17          IF (m.find (SUM) == m.end ()) // accumulation and absence of 
18 is              m [SUM] = I;
 . 19          Auto m.find PTR = (SUM - AIM);
 20 is          IF ! (PTR = m.end ()) // record the presence of 
21 is              RES = RES > (I - (ptr-> SECOND)) RES:? (I - (ptr-> SECOND)); // update the longest recording subarray 
22 is      }
 23 is      return res;
24 }
25 
26 void Test()
27 {
28     vector<int>v;
29     v = { 7,3,2,1,1,7,7,7 };
30     cout << mostLongSubArray(v, 7) << endl;
31     v = { 7,3,2,1,1,7,-6,-1,7 };
32     cout << mostLongSubArray(v, 7) << endl;
33     v = { 7,5,2,4,3,1,1,1,1,1,1,1,1,7 };
34     cout << mostLongSubArray(v, 7) << endl;
35 }