寻找连续子数组-优快云博客

http://blog.youkuaiyun.com/yysdsyl/article/details/5421200

题目：

You have given two arrays, say

A: 4, 1, 6, 2, 8, 9, 5, 3, 2, 9, 8, 4, 6
B: 6, 1, 2, 9, 8

where B contains elements which are in A in consecutive locations but may be in any order.
Find their starting and ending indexes in A. (Be careful of duplicate numbers).

answer is （1，5）

先给出代码，再结合代码解释，如下：

[cpp]view plaincopy 
   
 #include <iostream>  
 #include <map>  
   
 using namespace std;  
   
 void FindConsecutiveSubarrLocation(int A[], int lenA, int B[], int lenB)  
 {  
     map<int, int> bmap;  
     map<int, int> windowmap;  
     map<int, int> diffmap;  
     for(int i = 0; i< lenB; i++)  
     {  
         if(bmap.count(B[i]) == 0)  
             bmap[B[i]] = 1;  
         else  
             ++bmap[B[i]];  
         if(windowmap.count(A[i]) == 0)  
             windowmap[A[i]] = 1;  
         else  
             ++windowmap[A[i]];  
     }  
   
     map<int, int>::iterator it = bmap.begin();  
     int sameElement = 0;  
     while(it != bmap.end())  
     {  
         if(windowmap.count((*it).first) != 0)  
         {  
             diffmap[(*it).first] = windowmap[(*it).first] - (*it).second;  
             if(diffmap[(*it).first]  == 0)  
                 sameElement++;  
         }  
         else  
             diffmap[(*it).first] = -(*it).second;;  
         it++;  
     }  
   
     if(sameElement == lenB)  
     {  
         cout<<"----------find one---------"<<endl;  
         cout<<"start index:"<<0<<endl;  
         cout<<"end index:"<<lenB-1<<endl;  
     }  
   
     for(int i = lenB; i < lenA; i++)  
     {  
         if(diffmap.count(A[i-lenB]) != 0)  
         {  
             diffmap[A[i-lenB]]--;  
             if(diffmap[A[i-lenB]] == 0)  
                 sameElement++;  
             else if(diffmap[A[i-lenB]] == -1)  
                 sameElement--;  
         }  
         if(diffmap.count(A[i]) !=0)  
         {  
             diffmap[A[i]]++;  
             if(diffmap[A[i]] == 0)  
                 sameElement++;  
             else if(diffmap[A[i]] == 1)  
                 sameElement--;  
         }  
         if(sameElement == diffmap.size())  
         {  
             cout<<"----------find one---------"<<endl;  
             cout<<"start index:"<<i-lenB+1<<endl;  
             cout<<"end index:"<<i+1<<endl;  
         }  
     }  
 }  
   
 int main()  
 {  
     int A[] = {4, 1, 2, 1, 8, 9, 2, 1, 2, 9, 8, 4, 6};  
     int B[] = {1, 1, 2, 8, 9};  
     int lenA = sizeof(A)/sizeof(int);  
     int lenB = sizeof(B)/sizeof(int);  
     FindConsecutiveSubarrLocation(A, lenA, B, lenB);  
     return 0;  
 }  

思路：

ex，

int A[] = {4, 1, 2, 1, 8, 9, 5, 3, 2, 9, 8, 4, 6};
int B[] = {1, 1, 2, 8, 9};
int lenA = 13;
int lenB = 5;
map<int,int> bmap;
map<int,int> windowmap;
map<int,int> diffmap;

首先初始化map:
bmap = {1:2, 2:1, 8:1,9:1}
    windowmap = {1:2, 2:1, 4:1,8:1}
    diffmap = {1:0, 2:0, 8:0, 9:-1}
    其中"1: 0" 意味着数组A的当前滑动窗口正好和数组B包含相同数量的"1"，而 " 9:-1" 则表示A当前的滑动窗口和B比较缺少1个"9"。并且我们注意到diffmap和bmap拥有相同的key

代码中的变量"sameElement"代表的是diffmap中有多少pair与bmap匹配, 显然，初始化后的“sameElement”变量值会是3 ( 1:0, 2:0, 8:0 in diffmap).

接下来
在数组A中滑动size为lenB的窗口，没向前滑动一步，只需check滑动窗口左侧划出的元素El和右侧滑入的元素Er，更新diffmap和sameElement，如下：

       if(diffmap.count(A[i-lenB]) != 0)
        {
            diffmap[A[i-lenB]]--;
            if(diffmap[A[i-lenB]] == 0)
                sameElement++;
            else if(diffmap[A[i-lenB]] == -1)
                sameElement--;
        }
        if(diffmap.count(A[i]) !=0)
        {
            diffmap[A[i]]++;
            if(diffmap[A[i]] == 0)
                sameElement++;
            else if(diffmap[A[i]] == 1)
                sameElement--;
        }
        if(sameElement == diffmap.size())
        {
            cout<<"----------find one---------"<<endl;
            cout<<"start index:"<<i-lenB+1<<endl;
            cout<<"end index:"<<i+1<<endl;
        }